Circle 4 Question 15
15. 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… .
$(1988,2 M)$
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Answer:
Correct Answer: 15. $-\frac{9}{5}, \frac{12}{5}$ and $\frac{9}{5},-\frac{12}{5}$
Solution:
- Given, $C _1: x^{2}+y^{2}=16$
and let $C _2:(x-h)^{2}+(y-k)^{2}=25$
$\therefore$ Equation of common chords is $S _1-S _2=0$
$\therefore \quad 2 h x+2 k y=\left(h^{2}+k^{2}-9\right)$
$\therefore$ Its slope $=-\frac{h}{k}=\frac{3}{4}$
[given]
If $p$ be the length of perpendicular on it from the centre $(0,0)$ of $C _1$ of radius 4 , then $p=\frac{h^{2}+k^{2}-9}{\sqrt{4 h^{2}+4 k^{2}}}$.
Also, the length of the chord is
$$ 2 \sqrt{r^{2}-p^{2}}=2 \sqrt{4^{2}-p^{2}} $$
The chord will be of maximum length, if $\varphi=0$ or $h^{2}+k^{2}-9=0 \Rightarrow h^{2}+\frac{16}{9} h^{2}=9$
$$ \begin{aligned} \Rightarrow & & h & = \pm \frac{9}{5} \\ \therefore & & k & =\mp \frac{12}{5} \end{aligned} $$
Hence, centres are $\frac{9}{5},-\frac{12}{5}$ and $-\frac{9}{5}, \frac{12}{5}$
