Radius, perpendicularity and power: a connected method for circle equations and tangents
JEE Main · JEE Advanced80 min study guide10 indexed sections
BY THE END, YOU CAN
Recover a circle center and radius and recognize degenerate or empty equations.
Use center-to-line distance for secants, tangents and chord lengths.
Derive tangent, contact-chord and point-power formulas with their geometric conditions.
Analyze two-circle intersections, radical axes and circle families.
BEFORE YOU START
Distance and perpendicularity in the coordinate plane.
Quadratic equations, completing squares and line equations.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A circle is a fixed-distance locus
A circle with center C=(h,k) and radius r>0 is the set of points X=(x,y) satisfying |X−C|=r. Squaring gives (x−h)²+(y−k)²=r². The distance definition supplies the geometry; the equation permits algebraic intersection and tangent calculations.
The normalized general form is x²+y²+2gx+2fy+c=0. Completing squares gives center (−g,−f) and radius squared g²+f²−c. A positive value gives a genuine circle, zero gives a single point and a negative value gives no real points.
If the x² and y² coefficients are the same nonzero number and there is no xy term, divide by that common coefficient before reading the normalized parameters. Unequal quadratic coefficients generally describe another conic rather than a circle in the given equal-scale Cartesian axes.
A point lies outside, on or inside according as |P−C|²−r² is positive, zero or negative. This signed quantity is the power of P with respect to the circle. It is algebraically useful even for interior points, where real tangent segments do not exist.
ORIGINAL WORKED EXAMPLE
Complete the square before reading the geometry
Find the center and radius of x²+y²−6x+4y−12=0 and classify P=(6,2).
Group and complete squares: (x−3)²−9+(y+2)²−4−12=0.
Thus (x−3)²+(y+2)²=25, giving C=(3,−2) and r=5.
The displacement from C to P is (3,4), whose squared length is 9+16=25.
So P lies on the circle and its power is zero. The signs in the center are opposite those in the linear coefficients.
Answer Center (3,−2), radius 5; P=(6,2) lies on the circle.
02 / UNDERSTAND & APPLY
2. Geometric data determine the unknown coefficients
A center and radius determine the standard equation immediately. Three noncollinear points determine a unique circle: substitute them into x²+y²+Dx+Ey+F=0 and solve the three linear equations for D,E,F. Noncollinearity is essential; three distinct collinear points cannot lie on a genuine finite circle.
If A and B are endpoints of a diameter, a point X lies on the circle exactly when (X−A)·(X−B)=0. This is the right-angle condition at X, with the endpoints included algebraically. The center is the midpoint and the radius is half |AB|.
A circle tangent to a specified line has radius equal to the perpendicular distance from its center to that line. If its center is also constrained to another locus, combine these conditions. Squaring may be convenient, but radii are nonnegative and a genuine circle requires positive radius.
Whenever data produce more than one possible center, retain all configurations allowed by the wording. A circle tangent to both coordinate axes can have centers in several quadrants; restricting the center to the first quadrant is an extra condition that must come from the problem.
ORIGINAL WORKED EXAMPLE
A diameter gives both an equation and a check
Find the circle whose diameter endpoints are A=(−2,1) and B=(4,5).
The center is their midpoint C=(1,3). The squared diameter length is 6²+4²=52, so r²=52/4=13.
The standard equation is (x−1)²+(y−3)²=13.
Alternatively use (x+2)(x−4)+(y−1)(y−5)=0. Expanding gives x²+y²−2x−6y−3=0.
Completing the squares of the expanded equation returns the same standard form, and both endpoints satisfy it.
Answer (x−1)²+(y−3)²=13.
03 / UNDERSTAND & APPLY
3. The perpendicular foot organizes every line intersection
Let a line have perpendicular distance d from the circle center. If d>r it misses the circle; if d=r it is tangent; if d<r it cuts a chord with two distinct endpoints. This follows by splitting the radius to an endpoint into perpendicular components d and the half-chord length.
If H is the perpendicular foot from C to the line and u is a unit direction along the line, the intersection points are H±√(r²−d²)u when d≤r. The chord length is 2√(r²−d²). At tangency the square root vanishes and both points merge into H.
Algebraically, substituting a line parametrization into the circle equation gives a quadratic in the parameter. A positive discriminant means two intersections, zero means tangency and negative means no real intersection. The distance and discriminant methods encode the same geometric test.
Using a parametrized line or a normal-vector equation avoids losing vertical lines. A slope equation y=mx+b cannot represent them, so a calculation that searches over finite m alone may miss a valid vertical tangent.
ORIGINAL WORKED EXAMPLE
Chord length from the center-to-line distance
Find the chord cut by x=3 on x²+y²=25, including its midpoint and length.
The center is the origin and radius is 5. The line x=3 is distance 3 from the center, so it is a secant.
The perpendicular foot is H=(3,0), which is the chord midpoint.
The half-length is √(25−9)=4. Along the vertical line the endpoints are (3,4) and (3,−4).
The full chord length is 8. Substitution gives y²=16, independently confirming the endpoints.
4. A tangent is perpendicular to the contact radius
At a point T on a genuine circle, the tangent is the line through T perpendicular to CT. In vector form it is (T−C)·(X−T)=0. Because |T−C|²=r², this can also be written (T−C)·(X−C)=r².
For x²+y²=r² and a contact point (x₁,y₁) on the circle, the equation is xx₁+yy₁=r². For a translated circle, use the centered coordinates: (x₁−h)(x−h)+(y₁−k)(y−k)=r².
The contact point must lie on the circle. Inserting an arbitrary off-circle point into the tangent-at-a-point expression does not produce a tangent at that point. For an exterior point it instead gives the chord of contact of its two tangents, a different object discussed below.
For normalized S=x²+y²+2gx+2fy+c=0, the tangent at an on-circle point (x₁,y₁) is xx₁+yy₁+g(x+x₁)+f(y+y₁)+c=0. Derive this from the centered form rather than memorizing a symbol-replacement rule without its contact condition.
ORIGINAL WORKED EXAMPLE
Use the radius as a normal vector
Find the tangent to (x−3)²+(y+2)²=25 at T=(6,2).
The center is C=(3,−2), so the contact radius vector is T−C=(3,4). Its length is 5, confirming T lies on the circle.
The tangent through T with normal (3,4) is 3(x−6)+4(y−2)=0.
Simplify to 3x+4y−26=0.
The distance from C to this line is |9−8−26|/5=25/5=5, equal to the radius, independently checking tangency.
Answer The tangent is 3x+4y−26=0.
05 / UNDERSTAND & APPLY
5. Exterior points have two tangent contacts
If P lies outside a circle, the right triangle CTP gives PT²=CP²−r². The two tangent segments from P therefore have equal length. If P is on the circle there is one tangent line; if P is inside, no real tangent line through P exists.
To find tangent lines from P, take a general line through P and impose center-to-line distance r, or impose a repeated root in the line-circle intersection. Check a vertical line separately if using slope form. Once a line is found, its contact point is the perpendicular foot from C.
For each contact T, perpendicularity PT⊥CT gives (P−T)·(T−C)=0. Rearranging with |T−C|²=r² yields (P−C)·(T−C)=r². Thus both contact points lie on the chord-of-contact line (P−C)·(X−C)=r².
The chord of contact joins the two touching points; it is not one of the tangent lines. Its midpoint lies on CP, and its distance from C is r²/CP. The same algebraic line can be defined for other P as a polar, but its interpretation as a real contact chord requires P outside.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
For x²+y²=r², the chord of contact from (u,v) lies on ux+vy=r².
ORIGINAL WORKED EXAMPLE
Find both tangents and their contact chord
For x²+y²=25 and P=(10,0), find tangent length, both tangent lines and the contact points.
CP=10>5, so two tangents exist and each segment length is √(100−25)=5√3.
A nonvertical line through P is y=m(x−10). Its distance from the origin is 10|m|/√(1+m²). Equating this to 5 gives m²=1/3. The vertical line x=10 is distance 10 from the center and is not tangent.
Thus the tangent lines are y=±(x−10)/√3. The contact chord is 10x=25, or x=5/2.
Intersecting x=5/2 with the circle gives y=±5√3/2. Each contact lies on one of the two lines; the positive contact belongs to the negative-slope tangent.
Parametrize a line through P as X=P+tu with a unit direction u. Substituting into |X−C|²=r² gives t²+2u·(P−C)t+(|P−C|²−r²)=0. By the quadratic root product, t₁t₂ equals the power CP²−r².
The parameters are directed distances along u. For an exterior secant whose two intersections lie on the same ray, the ordinary positive segment lengths multiply to the positive power. For P inside, the two intersections lie in opposite directions and the directed product is negative; the product of their ordinary lengths is the negative of the power.
At tangency the two parameter roots coincide, making the product the square of the tangent length. This recovers the right-triangle formula and explains why tangent and secant products are the same invariant.
The unit-direction condition matters in the parameter derivation. If the direction vector is not unit length, the leading quadratic coefficient is |u|², and the parameter product is power divided by |u|². Parameters and physical distances are then not numerically identical.
ORIGINAL WORKED EXAMPLE
Check a secant product against a tangent length
An exterior point is distance 13 from the center of a circle of radius 5. A secant meets the nearer point at distance 4 from it. Find the farther distance and tangent length.
The power is 13²−5²=169−25=144.
The two intersections are on the same ray, so their positive distances have product 144. If the nearer distance is 4, the farther distance is 144/4=36.
The tangent length is √144=12.
However, a circle of radius 5 has every chord length at most 10. The implied chord length would be 36−4=32, which is impossible. Therefore the supplied nearer-secant distance 4 is inconsistent with the stated circle geometry; no such secant exists.
Answer The data are inconsistent: the product would force distance 36 and an impossible chord length 32. The exterior point itself has tangent length 12.
07 / UNDERSTAND & APPLY
7. Subtract normalized circle equations to compare powers
For two circles with distinct centers, subtract their normalized equations S₁−S₂=0. The x²+y² terms cancel, leaving the radical axis: the line of points with equal powers. It is perpendicular to the line joining the centers because its normal is parallel to their center displacement.
If the circles intersect, their intersection points have power zero to both, so the radical axis is their common chord line. If they are tangent, it is the common tangent at the touching point. If they do not meet, the radical axis can still be a real line even though there is no real common chord.
Distinct concentric circles have no points of equal power because the difference of their powers is a nonzero constant. Identical circles have equal power everywhere. These cases do not produce a unique radical axis line and must be separated from the distinct-center theorem.
At a point on the radical axis that lies outside both circles, the two tangent lengths are equal because their squares are equal powers. On an interior portion, power may be negative and there are no real tangent lengths to equate. Equal power is the general statement.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Subtracting normalized circle equations cancels x²+y².
ORIGINAL WORKED EXAMPLE
A common chord and an exterior equal-power point
For S₁:x²+y²=25 and S₂:(x−4)²+y²=9, find their common chord and the tangent lengths from P=(4,5).
Normalize S₂ to x²+y²−8x+7=0, while S₁ is x²+y²−25=0. Subtracting gives 8x−32=0, hence x=4.
On S₁, x=4 gives y²=9, so the intersection points are (4,3) and (4,−3). Their common chord length is 6.
P=(4,5) lies on the radical axis. Its power to S₁ is 4²+5²−25=16.
Its power to S₂ is (4−4)²+5²−9=16. Both are positive, so the tangent lengths to both circles are √16=4.
Answer Common chord x=4 with endpoints (4,±3); both tangent lengths from P are 4.
08 / UNDERSTAND & APPLY
8. Radius comparisons and circle families need exceptional cases
For radii R,r>0 and center separation d>0, the circles intersect in two points when |R−r|<d<R+r. They touch externally at d=R+r and internally at d=|R−r|. If d>R+r they are separate; if d<|R−r| one lies strictly inside the other without touching.
For distinct nonconcentric circles, the number of common tangent lines is four in the separate external case, three at external tangency, two when they intersect, one at internal tangency and zero when one is strictly inside the other. Identical circles have infinitely many common tangents and are a separate case.
A family through two fixed circle intersections can be written S₁+λ(S₁−S₂)=0, where both S expressions are normalized. The squared coefficients remain equal to 1, so each finite λ gives a circle equation through the common points. For two distinct real common points these members are genuine circles.
The alternative homogeneous form αS₁+βS₂=0 includes a line when α+β=0: that member is the radical axis rather than a circle. A parameterized family should therefore be classified after simplification, especially if the quadratic coefficient can vanish.
09 / UNDERSTAND & APPLY
9. Original practice with full geometric checks
Confirm that contact points lie on the circle and that proposed lengths fit the geometry. All circles below have positive radius unless a question asks about degeneracy.
ORIGINAL PRACTICE · 1
Find center and radius of x²+y²+4x−6y−3=0.
Reveal answer & reasoning
Center (−2,3), radius 4.
Completing squares gives (x+2)²+(y−3)²=16.
ORIGINAL PRACTICE · 2
What real locus is x²+y²−2x+4y+5=0?
Reveal answer & reasoning
The single point (1,−2).
Completing squares gives (x−1)²+(y+2)²=0, a radius-zero degenerate circle equation.
ORIGINAL PRACTICE · 3
Find the tangent to x²+y²=10 at (1,3).
Reveal answer & reasoning
x+3y=10.
The point lies on the circle since 1+9=10. Its radius vector (1,3) is the tangent normal.
ORIGINAL PRACTICE · 4
Find the length of the chord y=4 in x²+y²=25.
Reveal answer & reasoning
6.
The center-to-line distance is 4, so the half-chord length is √(25−16)=3.
ORIGINAL PRACTICE · 5
Find tangent length from (8,0) to x²+y²=16.
Reveal answer & reasoning
4√3.
Power is 64−16=48, and tangent length is √48.
ORIGINAL PRACTICE · 6
How many tangents to x²+y²=9 pass through (1,1)?
Reveal answer & reasoning
None.
The point is inside because 1²+1²=2<9. Its negative power cannot be a real tangent-length square.
ORIGINAL PRACTICE · 7
Find the contact chord from P=(0,10) to x²+y²=25.
Reveal answer & reasoning
y=5/2.
The formula P·X=r² gives 10y=25. This is the line joining contacts, not a tangent through P.
ORIGINAL PRACTICE · 8
Find the radical axis of x²+y²=4 and (x−3)²+y²=1.
Reveal answer & reasoning
x=2.
The normalized second equation is x²+y²−6x+8=0. Subtracting the first gives −6x+12=0.
ORIGINAL PRACTICE · 9
Two circles have radii 5 and 2 and center distance 8. How many common tangents exist?
Reveal answer & reasoning
Four.
Since 8>5+2, the circles are separate externally, admitting two direct and two transverse common tangents.
ORIGINAL PRACTICE · 10
Two circles have radii 5 and 2 and center distance 3. What is their relative position?
Reveal answer & reasoning
Internally tangent, with one common tangent.
The separation equals the radius difference |5−2|=3.
ORIGINAL PRACTICE · 11
For a circle of radius 5, an exterior point has center distance 13. Its nearer secant distance is 8. Find the farther distance.
Reveal answer & reasoning
18.
Power is 144, so the farther distance is 144/8=18. The chord length is 10, consistent with a diameter secant through the center.
ORIGINAL PRACTICE · 12
Do two distinct concentric circles have a radical axis line?
Reveal answer & reasoning
No.
Their power difference is a nonzero constant at every point, so the equal-power locus is empty.
10 / UNDERSTAND & APPLY
10. Normalize, identify the geometry and verify
Begin by putting the circle in normalized or centered form. Then decide whether the problem concerns a point position, a line intersection, a tangent contact, a power relation or a comparison of two circles. Each has a direct geometric invariant: radius, perpendicular distance or power.
The common errors are reading the center with the wrong signs, treating a radius-squared value as the radius, using the tangent-at-contact formula for an off-circle point, forgetting vertical tangents in a slope search, and confusing a contact chord with either tangent line.
For secant products, distinguish directed parameters from ordinary lengths and check whether the point is inside or outside. For radical axes, normalize both equations and separate concentric cases. For families, check whether the quadratic terms cancel.
Finish by checking existence as well as algebra. A chord cannot exceed the diameter, a tangent contact must lie on the circle, and the radius to that contact must be perpendicular to the tangent. These geometric checks can reveal inconsistent data even when a product equation has a numerical solution.
KEEP THE BIG IDEAS
Quick revision
A centered circle equation exposes radius and point power.
Center-to-line distance classifies secants and tangents and gives chord length.
A tangent is perpendicular to the radius at an on-circle contact point.
Power links directed secant products and exterior tangent-length squares.
The radical axis compares normalized powers and needs separate concentric cases.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
Instructor-authored reference inspected for point power, secant signs and equal-power loci. The lesson explicitly restricts tangent-length interpretations to exterior points and separates concentric circles.