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If the circle \mathrm{C_1 \equiv x^2+y^2=16}  intersects another circle  \mathrm{C_2} of radius 5 in such a manner that the common chord is of maximum length and has a slope equal to \mathrm{\frac{3}{4}} then the coordinates of centre \mathrm{C_{2}} are

Option: 1

\mathrm{\left(\frac{9}{5},-\frac{12}{5}\right) or \left(-\frac{9}{5}, \frac{12}{5}\right)}


Option: 2

\mathrm{\left(\frac{5}{4}, \frac{-7}{5}\right) or \left(\frac{-5}{4}, \frac{7}{5}\right)}


Option: 3

\mathrm{\left(\frac{9}{5}, \frac{12}{5}\right) or \left(-\frac{9}{5},-\frac{12}{5}\right)}


Option: 4

\mathrm{\left(-\frac{5}{4}, \frac{-7}{5}\right) or \left(\frac{5}{4}, \frac{7}{5}\right)}


Answers (1)

best_answer

Let (h, k) be the coordinates of centre of circle \mathrm{C}_2

Its equation is then

\mathrm{(\mathrm{x}-\mathrm{h})^2+(\mathrm{y}-\mathrm{k})^2=25}----(1)

The equation of common chord is

\mathrm{\begin{aligned} & \left(x^2+y^2-16\right)-\left[(x-h)^2+(y-k)^2-25\right]=0 \\ & \Rightarrow 2 h x+2 k y-h^2-k^2+9=0 \end{aligned}}--------(2)

Its slope \mathrm{=-\frac{\mathrm{h}}{\mathrm{k}}=\frac{3}{4} \text { (given) }}

\mathrm{\Rightarrow \mathrm{k}=-\frac{4 h}{3}} ______(3)

Length of perpendicular from centre (0, 0) of given circle on the chord (2)

\mathrm{\mathrm{p}=\frac{h^2+k^2-9}{2 \sqrt{h^2+k^2}}}

The length of common chord

\mathrm{=2 \sqrt{(\text { radius })^2-p^2}=2 \sqrt{16-p^2}}

For maximum length of the chord, p = 0

\mathrm{\begin{aligned} & \Rightarrow \frac{h^2+k^2-9}{2 \sqrt{h^2+k^2}}=0 \quad \Rightarrow h^2+\mathrm{k}^2=9 \\ & \Rightarrow \mathrm{h}^2+\frac{16 h^2}{9}=9 \quad \Rightarrow \mathrm{h}= \pm \frac{\frac{9}{}}{5} \\ & \end{aligned}}

And then \mathrm{\mathrm{k}=\square \frac{12}{5}}

Thus centre of  \mathrm{\mathrm{C}_2} is \mathrm{\left(\frac{9}{5},-\frac{12}{5}\right) \text { or }\left(-\frac{9}{5}, \frac{12}{5}\right)}

 

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