Question

Circle $C_1:x^2+y^2=16$ intersects another circle $C_2$ of radius $6$ in such a manner that their common chord is of maximum length and has slope equal to $\frac{1}{2}.$ Then the co-ordinates of the centre of the circle(s) $C_2$ is (are)

• A. $(-2,-4)$
• B. $(2,4)$
• C. $(-2,4)$
• D. $(2,-4)$

Answer 1

Answer: (C), (D)

$(2,-4)$

According to the question, the chord is of maximum possible length
$\Rightarrow$ The diameter of the circle $C_1:$$x^2+y^2=16$ is our required common chord .
$\because$ Slope of the chord is given as $\frac{1}{2},$
$\Rightarrow$ Slope of line perpendicular to the chord would be $-2$ and its equation would be $L:2x+y=c$
$\because L$ will pass through both of the centres of the circles.
$\Rightarrow L:2x+y=0$
Let the centre of the circle $C_2$ be $Q(h,k)$
$\Rightarrow 2h+k=0...\left(1\right)$

Now, In $\Delta PAQ$
$6^2=4^2+A{Q}^2\text{and}A{Q}^2=h^2+k^2\Rightarrow h^2+k^2=20...\left(2\right)$
Using $\left(1\right)$ and $\left(2\right),$ we get
$h=\pm 2\Rightarrow k=\mp 4$
$\therefore$ The co-ordinates of the circle $C_2$ are $(2,-4)$ and $(-2,4)$.