If the circle {C_1}: {x^2} + {y^2} = 16 inter | vedclass

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(A) Let $(h, k)$ be the coordinates of the centre of circle ${C_2}$.Its equation is $(x - h)^2 + (y - k)^2 = 5^2$.The equation of ${C_1}$ is $x^2 + y^

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

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(A) Let $(h, k)$ be the coordinates of the centre of circle ${C_2}$.Its equation is $(x - h)^2 + (y - k)^2 = 5^2$.The equation of ${C_1}$ is $x^2 + y^2 = 4^2$. The equation of the common chord of ${C_1}$ and ${C_2}$ is $2hx + 2ky = h^2 + k^2 - 9$ $(i)$.Let $p$ be the length of the perpendicular from the centre $(0, 0)$ of ${C_1}$ to $(i)$.Then $p = \frac{|h^2 + k^2 - 9|}{\sqrt{4h^2 + 4k^2}}$.The length of the common chord is $2\sqrt{4^2 - p^2}$,which is of maximum length if $p = 0$.This implies $h^2 + k^2 - 9 = 0$ $(ii)$.The slope of $(i)$ is given as $\frac{3}{4}$.Thus,$-\frac{h}{k} = \frac{3}{4} \Rightarrow k = -\frac{4h}{3}$ $(iii)$.Substituting $(iii)$ into $(ii)$,we get $h^2 + (-\frac{4h}{3})^2 = 9$ $\Rightarrow h^2 + \frac{16h^2}{9} = 9$ $\Rightarrow \frac{25h^2}{9} = 9$ $\Rightarrow h^2 = \frac{81}{25}$ $\Rightarrow h = \pm \frac{9}{5}$.If $h = \frac{9}{5}$,then $k = -\frac{4}{3} 	imes \frac{9}{5} = -\frac{12}{5}$.If $h = -\frac{9}{5}$,then $k = -\frac{4}{3} 	imes (-\frac{9}{5}) = \frac{12}{5}$.Thus,the centres are $\left( -\frac{9}{5}, \frac{12}{5} ight)$ and $\left( \frac{9}{5}, -\frac{12}{5} ight)$.

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 $\frac{3}{4}$,the coordinates of the centre of ${C_2}$ are

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