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Question
If chords are drawn from the point (2,0)(2, 0) to the circle x2+y2=4x^2 + y^2 = 4, then the locus of the midpoint of each chord is:
(A)x2+y2=2xx^2 + y^2 = 2x
(B)x2+y2=4xx^2 + y^2 = 4x
(C)x2+y2=xx^2 + y^2 = x
(D)x2+y2=2yx^2 + y^2 = 2y
Solution Path
Chord with midpoint (h,k) has equation hx+ky = h^2+k^2 by T=S1. Passes through (2,0): 2h = h^2+k^2. Locus: x^2+y^2 = 2x, a circle with center (1,0) and radius 1.
01Question Setup
1/4
Chords are drawn from (2,0)(2,0) to the circle x2+y2=4x^2+y^2=4. Find the locus of the midpoint of each chord.
Locus =  ?= \;?
02Chord with Given Midpoint
2/4
Using T=S1T = S_1: the chord with midpoint (h,k)(h,k) has equation hx+ky=h2+k2hx + ky = h^2 + k^2. This chord passes through (2,0)(2,0).
hx+ky=h2+k2hx + ky = h^2 + k^2
03Apply Point ConditionKEY INSIGHT
3/4
Substituting (2,0)(2,0): 2h=h2+k22h = h^2 + k^2. Replace (h,k)(h,k) with (x,y)(x,y) for the locus equation.
x2+y2=2xx^2 + y^2 = 2x
04Final Answer
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Locus: x2+y2=2xx^2+y^2 = 2x, equivalently (x−1)2+y2=1(x-1)^2+y^2=1. This is a circle with center (1,0)(1,0) and radius 11.
x2+y2=2x\boxed{x^2+y^2=2x}
Concepts from this question2 concepts unlocked
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Power of a Point

EASY

The power of a point P(x1, y1) with respect to a circle S: x^2 + y^2 + 2gx + 2fy + c = 0 is defined as S1 = x1^2 + y1^2 + 2gx1 + 2fy1 + c. It equals (distance from center)^2 - r^2. If P is outside, S1 > 0 and equals the square of the tangent length. If P is on the circle, S1 = 0. If P is inside, S1 < 0.

S1=x12+y12+2gx1+2fy1+c=d2−r2S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c = d^2 - r^2

Power of a point unifies tangent length, position of a point, and radical axis into one concept. JEE frequently tests whether students can correctly determine if a point is inside, outside, or on a circle using S1.

Length of tangentPosition of point relative to circleRadical axisChord of contact
Practice (10 Qs) →

Tangent from an External Point

STANDARD

From a point outside a circle, exactly two tangent lines can be drawn. For the circle x^2 + y^2 = a^2, a tangent with slope m has the equation y = mx +/- a*sqrt(1 + m^2). The length of each tangent equals sqrt(S1). The chord of contact joining the two tangent points has equation T = 0.

y=mx±a1+m2,L=S1y = mx \pm a\sqrt{1 + m^2}, \quad L = \sqrt{S_1}

Tangent problems are the most frequently tested subtopic in Circles. JEE tests tangent equations, tangent lengths, and the number of common tangents between two circles. Missing the +/- sign is the most common error.

Tangent with given slopeTangent from external pointCommon tangents to two circlesDirector circle
Practice (12 Qs) →
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