Question

The radius of a circle $C_1$ is thrice the radius of another circle $C_2$ and the centres of $C_1$ and $C_2$ are (1,2) and (3,-2) respectively. If they cut each other orthogonally and the radius of the circle $C_1$ is 3r, then the equation of the circle with r as radius and (1,-2) as centre is

• A. $x^2+y^2-2x+4y-3=0$
• B. $x^2+y^2-2x+4y+7=0$
• C. $x^2+y^2-2x+4y-7=0$
• D. $x^2+y^2-2x+4y+3=0$

Hint:

The condition for orthogonality, $d^2 = r_1^2 + r_2^2$, is fundamental. It arises from applying the Pythagorean theorem to the triangle formed by the two centers and one of their intersection points, where the tangents at that point are perpendicular.

Answer 1

The Correct Option is D

Step 1: Find Radius r: Let radii be \( r_1 = 3r \) and \( r_2 = r \) (since question implies \( C_2 \) corresponds to r). Distance between centers \( d^2 = (3-1)^2 + (-2-2)^2 = 4 + 16 = 20 \). Orthogonality condition: \( d^2 = r_1^2 + r_2^2 \). \( 20 = (3r)^2 + r^2 = 10r^2 \). \( r^2 = 2 \).
Step 2: Equation of Required Circle: Centre given: \( (1, -2) \). Radius: \( r = \sqrt{2} \). \( (x-1)^2 + (y+2)^2 = (\sqrt{2})^2 \) \( x^2 - 2x + 1 + y^2 + 4y + 4 = 2 \) \( x^2 + y^2 - 2x + 4y + 3 = 0 \).