The point/points of intersection of the commo | vedclass

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(B) Given circles are:$S_1: x^2+y^2-8x-6y+21=0$$S_2: x^2+y^2-2y-15=0$For circle $S_1$,the center is $C_1(4,3)$ and the radius is $r_1 = \sqrt{4^2+3^2-

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$(8,5)$

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(B) Given circles are:$S_1: x^2+y^2-8x-6y+21=0$$S_2: x^2+y^2-2y-15=0$For circle $S_1$,the center is $C_1(4,3)$ and the radius is $r_1 = \sqrt{4^2+3^2-21} = \sqrt{16+9-21} = \sqrt{4} = 2$.For circle $S_2$,the center is $C_2(0,1)$ and the radius is $r_2 = \sqrt{0^2+1^2-(-15)} = \sqrt{1+15} = \sqrt{16} = 4$.The distance between the centers is $C_1C_2 = \sqrt{(4-0)^2+(3-1)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.Since $|r_1-r_2| The external center of similitude $P$ divides the line segment joining the centers $C_1$ and $C_2$ externally in the ratio of their radii $r_1:r_2 = 2:4 = 1:2$.Using the external division formula for $C_1(4,3)$ and $C_2(0,1)$ with ratio $1:2$:$P = \left( \frac{m_1x_2 - m_2x_1}{m_1-m_2}, \frac{m_1y_2 - m_2y_1}{m_1-m_2} \right) = \left( \frac{1(0) - 2(4)}{1-2}, \frac{1(1) - 2(3)}{1-2} \right)$$P = \left( \frac{-8}{-1}, \frac{-5}{-1} \right) = (8,5)$.

The point/points of intersection of the common tangents of the two circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2y-15=0$ is/are

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