The tangent to the circle `x^2 + y^2 = 5` at (1, -2) also touches the circle `x^2 +y^2- 8x + 6y + 20=0`. Find the coordinates of the corresponding point of contact.

The tangent to the circle `x^2 + y^2 = 5` at (1, -2) also touches the

1.	The tangent to the circle `x^2 + y^2 = 5` at (1, -2) also touches the circle `x^2 +y^2- 8x + 6y + 20=0`. Find the coordinates of the corresponding point of contact.
Answer» Equation of tangent to `x^(2)+y^(2)=5` at (1,-2) is `x-2y-5=0`. Solving this with the second circle, we get `(2y+5)^(2)+y^(2)-8(2y+5)+6y+20=0` `implies 5y^(2)+10y+5=0` `implies (y+1)^(2)=0` `implies y=-1` `implies x =-2+5=3` Thus, point of contact on second circle is `(3,-1)`.

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The tangent to the circle `x^2 + y^2 = 5` at (1, -2) also touches the circle `x^2 +y^2- 8x + 6y + 20=0`. Find the coordinates of the corresponding point of contact.

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