JEE PYQ: Circle Question 17
Question 17 - 2020 (04 Sep Shift 2)
The circle passing through the intersection of the circles, $x^2 + y^2 - 6x = 0$ and $x^2 + y^2 - 4y = 0$, having its centre on the line, $2x - 3y + 12 = 0$, also passes through the point:
(1) $(-1, 3)$
(2) $(-3, 6)$
(3) $(-3, 1)$
(4) $(1, -3)$
Show Answer
Answer: (2)
Solution
Family: $S_1 + \lambda S_2 = 0$. $(1+\lambda)x^2 + (1+\lambda)y^2 - 6x - 4\lambda y = 0$. Center $\left(\frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda}\right)$ on $2x - 3y + 12 = 0$: $\frac{6}{1+\lambda} - \frac{6\lambda}{1+\lambda} + 12 = 0 \Rightarrow \lambda = -3$. Circle: $x^2 + y^2 + 3x - 6y = 0$. $(-3,6)$ satisfies: $9 + 36 - 9 - 36 = 0$.
BETA
