The point of intersection of the common tange | vedclass

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(C) Given equations of the circles are:$x^2+y^2-4x-2y+1=0 \quad \dots (i)$$x^2+y^2-6x-4y+4=0 \quad \dots (ii)$For circle $(i)$,center $C_1 = (2, 1)$ a

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$(0, -1)$

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(C) Given equations of the circles are:$x^2+y^2-4x-2y+1=0 \quad \dots (i)$$x^2+y^2-6x-4y+4=0 \quad \dots (ii)$For circle $(i)$,center $C_1 = (2, 1)$ and radius $r_1 = \sqrt{2^2+1^2-1} = 2$.For circle $(ii)$,center $C_2 = (3, 2)$ and radius $r_2 = \sqrt{3^2+2^2-4} = 3$.The distance between centers $C_1C_2 = \sqrt{(3-2)^2+(2-1)^2} = \sqrt{1^2+1^2} = \sqrt{2}$.Since $|r_1-r_2| The point of intersection of external common tangents divides the line joining the centers externally in the ratio $r_1:r_2$.Let the point be $P(x, y)$. Using the external division formula:$x = \frac{r_1x_2 - r_2x_1}{r_1-r_2} = \frac{2(3) - 3(2)}{2-3} = \frac{6-6}{-1} = 0$$y = \frac{r_1y_2 - r_2y_1}{r_1-r_2} = \frac{2(2) - 3(1)}{2-3} = \frac{4-3}{-1} = -1$Thus,the point of intersection is $(0, -1)$.

The point of intersection of the common tangents drawn to the circles $x^2+y^2-4x-2y+1=0$ and $x^2+y^2-6x-4y+4=0$ is:

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