CircleHardBloom L3

Question

If the circle $C_{1}:x^{2} + y^{2} = 16$ intersects another circle $C_{2}$ of radius 5 in such a manner that the common chord is of maximum length and has a slope equal to $3/4$, then the coordinates of the centre of $C_{2}$ are:

Options

A.$\left( \pm \frac{9}{5}, \pm \frac{12}{5} \right)$
B.$\left( \pm \frac{9}{5}, \mp \frac{12}{5} \right)$
C.$\left( \pm \frac{12}{5}, \pm \frac{9}{5} \right)$
D.$\left( \pm \frac{12}{5}, \mp \frac{9}{5} \right)$

Solution

[Key Concept: Two intersecting circles: the common chord is the radical axis; its length sets the condition.]

Slope of $C_{1}C_{2} = - \frac{4}{3} = tan\theta$

$$\Rightarrow C_{2} \equiv \left( 0 + 3\left( - \frac{3}{5} \right),0 + 3\left( \frac{4}{5} \right) \right) $$or $\left( 0 - 3\left( - \frac{3}{5} \right),0 - 3\left( \frac{4}{5} \right) \right)$

$$\begin{matrix} C_{2} & \ \equiv \left( \frac{9}{5}, - \frac{12}{5} \right)\text{~}\text{or}\text{~}\left( - \frac{9}{5},\frac{12}{5} \right) \\ 4r & \ = 12 \Rightarrow r = 3 \end{matrix}$$

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