A circle S equiv x^2+y^2-16=0 intersects anot | vedclass

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(C) The circle $S \equiv x^2+y^2=16$ has center $C_1(0,0)$ and radius $r_1=4$.For the common chord to be of maximum length,it must be the diameter of

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

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(C) The circle $S \equiv x^2+y^2=16$ has center $C_1(0,0)$ and radius $r_1=4$.For the common chord to be of maximum length,it must be the diameter of the smaller circle $S$. Thus,the common chord is a diameter of $S$,passing through the center $(0,0)$.The slope of the common chord is $m = \frac{3}{4}$. The equation of the line containing the chord is $y - 0 = \frac{3}{4}(x - 0)$,which simplifies to $3x - 4y = 0$.The center $C_2(h, k)$ of the circle $S^{\prime}$ must lie on the line perpendicular to the common chord passing through the center of $S$,because the line joining the centers of two intersecting circles is perpendicular to their common chord.The slope of the line joining the centers is $m_{\perp} = -\frac{1}{m} = -\frac{4}{3}$.The equation of the line of centers is $y - 0 = -\frac{4}{3}(x - 0)$,or $4x + 3y = 0$.Since the common chord is the diameter of $S$,the distance from $C_2$ to the chord is $d = \sqrt{r_2^2 - R_{chord}^2}$. Here $r_2 = 5$ and $R_{chord} = 4$,so $d = \sqrt{5^2 - 4^2} = 3$.The distance from $(h, k)$ to $3x - 4y = 0$ is $\frac{|3h - 4k|}{\sqrt{3^2 + (-4)^2}} = 3$,so $|3h - 4k| = 15$.Substituting $k = -\frac{4}{3}h$ into the equation: $|3h - 4(-\frac{4}{3}h)| = 15 \implies |3h + \frac{16}{3}h| = 15 \implies |\frac{25}{3}h| = 15 \implies |h| = \frac{45}{25} = \frac{9}{5}$.If $h = \frac{9}{5}$,then $k = -\frac{4}{3}(\frac{9}{5}) = -\frac{12}{5}$. If $h = -\frac{9}{5}$,then $k = \frac{12}{5}$.Checking the options,$\left(-\frac{9}{5}, \frac{12}{5}\right)$ is option $C$.

$A$ circle $S \equiv x^2+y^2-16=0$ intersects another circle $S^{\prime}=0$ of radius $5$ units such that their common chord is of maximum length. If the slope of that chord is $\frac{3}{4}$,then the centre of such a circle $S^{\prime}=0$ is

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