If the circle S_1: x^2+y^2=16 intersects anot | vedclass

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(A) Given $S_1: x^2+y^2=16$. The center is $B(0,0)$ and the radius is $r_1=4$. Since the common chord is of maximum length,it must be the diameter of

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

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(A) Given $S_1: x^2+y^2=16$. The center is $B(0,0)$ and the radius is $r_1=4$. Since the common chord is of maximum length,it must be the diameter of $S_1$. Let $PQ$ be the common chord. The length of the common chord is $2 	imes 4 = 8$. Let the center of $S_2$ be $A(h, k)$ and its radius be $r_2=5$. The distance from the center $A(h, k)$ to the common chord $PQ$ is $d = \sqrt{r_2^2 - (	ext{half-length of chord})^2} = \sqrt{5^2 - 4^2} = \sqrt{25-16} = 3$. The common chord $PQ$ passes through the origin $(0,0)$ and has a slope $m = \frac{3}{4}$. The line $AB$ is perpendicular to the common chord $PQ$. Thus,the slope of $AB$ is $m' = -\frac{1}{m} = -\frac{4}{3}$. The distance $AB = 3$. Using the parametric form of the line $AB$ passing through $(0,0)$ with slope $-\frac{4}{3}$: $h = 0 \pm 3 \cos 	heta$ and $k = 0 \pm 3 \sin 	heta$,where $	an 	heta = -\frac{4}{3}$. Since $	an 	heta = -\frac{4}{3}$,we have $\cos 	heta = \mp \frac{3}{5}$ and $\sin 	heta = \pm \frac{4}{3} 	imes \frac{3}{5} = \pm \frac{4}{5}$. Therefore,$(h, k) = \pm 3 \left(-\frac{3}{5}, \frac{4}{5}ight) = \left(\mp \frac{9}{5}, \pm \frac{12}{5}ight)$. The centers are $\left(-\frac{9}{5}, \frac{12}{5}ight)$ and $\left(\frac{9}{5}, -\frac{12}{5}ight)$.

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 has a slope of $\frac{3}{4}$,then the center of the circle $S_2$ is

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