Q. The other end of the diameter through the point $( - 1,\ 1)$ on the circle $x^{2} + y^{2} - 6x + 4y - 12 = 0$ is
A) $( - 7,\ 5)$
B) $( - 7,\ - 5)$
C) $(7,\ - 5)$
D) (7, 5)
Explanation
(c)
Given circle is $x^{2} + y^{2} - 6x + 4y - 12 = 0$
Centre of this circle is $(3,\ - 2)$
Let other end of the diameter is $(\alpha,\ \beta)$
$\therefore\ \ \ \frac{\alpha - 1}{2} = 3,\ \ \frac{\beta + 1}{2} = - 2$
$\Rightarrow \ \ \ \alpha = 7,\ \beta = - 5$
$\therefore$ Other end of the diameter is $(7,\ - 5)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.