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If the circle $S_1: x^2+y^2=16$ intersects another circle $S_2$ of radius 5 units such that the common chord is of maximum length and slope $\frac{3}{4}$, then the centre of the circle $S_2$ is
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Verified Answer
The correct answer is:
$\left(\frac{-9}{5}, \frac{12}{5}\right)$ or $\left(\frac{9}{5}, \frac{-12}{5}\right)$
Given $S_1=x^2+y^2=16$
Centre $(0,0) r=4$
Clearly, the diameter of $S_1$ will be common chord. Let $P Q$ be common chord and centre of $S_2$ be $(h, k)$
We have $A P=5, P B=4$,
$\therefore \quad A B=3$
Where $B=(0,0)$
Slope of $P Q=\frac{3}{4}$
$\therefore$ Slope of $A B=\frac{-4}{3}$
By parameter equation
$\frac{h-0}{\frac{-3}{5}}=\frac{k-0}{\frac{4}{5}}= \pm 3$
$x=\mp \frac{9}{5}, y=\frac{ \pm 12}{5}$
$\therefore$ Centre $\left(\frac{-9}{5}, \frac{12}{5}\right)$ and $\left(\frac{9}{5}, \frac{-12}{5}\right)$
Centre $(0,0) r=4$
Clearly, the diameter of $S_1$ will be common chord. Let $P Q$ be common chord and centre of $S_2$ be $(h, k)$
We have $A P=5, P B=4$,
$\therefore \quad A B=3$
Where $B=(0,0)$
Slope of $P Q=\frac{3}{4}$
$\therefore$ Slope of $A B=\frac{-4}{3}$
By parameter equation
$\frac{h-0}{\frac{-3}{5}}=\frac{k-0}{\frac{4}{5}}= \pm 3$
$x=\mp \frac{9}{5}, y=\frac{ \pm 12}{5}$
$\therefore$ Centre $\left(\frac{-9}{5}, \frac{12}{5}\right)$ and $\left(\frac{9}{5}, \frac{-12}{5}\right)$
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