Visual SolutionTricky

Question

The radical center of the circles

x^2 + y^2 = 4

x^2 + y^2 - 6x = 0

, and

x^2 + y^2 - 4y = 0

is:

(A)

\left(\dfrac{2}{3}, \dfrac{1}{2}\right)

(B)

\left(\dfrac{2}{3}, 1\right)

(C)

(1, 1)

(D)

(2, 2)

Solution Path

Radical axis of C1,C2 gives x=2/3. Radical axis of C1,C3 gives y=1. Radical center = (2/3, 1).

01Question Setup

1/4

Find the radical center of

C_1: x^2+y^2=4

C_2: x^2+y^2-6x=0

, and

C_3: x^2+y^2-4y=0

Radical center

= \;?

02Radical Axes

2/4

RA of

C_1, C_2

S_1 - S_2 = 6x - 4 = 0

, so

x = 2/3

. RA of

C_1, C_3

S_1 - S_3 = 4y - 4 = 0

, so

y = 1

x = 2/3, \; y = 1

03Find IntersectionKEY INSIGHT

3/4

The radical center is the intersection of any two radical axes. From

x = 2/3

and

y = 1

: radical center

= (2/3, 1)

(2/3, \; 1)

04Final Answer

4/4

Radical center

= (2/3, 1)

. This point has equal power with respect to all three circles.

\boxed{\left(\dfrac{2}{3}, 1\right)}

Concepts from this question1 concepts unlocked

★

Radical Axis of Two Circles

STANDARD

The radical axis of two circles S1 and S2 is the locus of points having equal power with respect to both circles. Its equation is S1 - S2 = 0 (after normalizing both equations so that the coefficients of x^2 and y^2 are 1). The radical axis is always a straight line perpendicular to the line joining the centers.

S_1 - S_2 = 0 \quad (\text{after normalizing})

The radical axis connects circle geometry to linear algebra. JEE tests it in combination with family of circles and common chords. The radical center (intersection of three pairwise radical axes) is another important concept.

Common chord of two circlesRadical centerCoaxial circlesEqual tangent length locus

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