The tangent to the circle x^2 + y^2 = 5 at th | vedclass

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(D) The equation of the tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ is given by $x(1) + y(-2) = 5$,which simplifies to $x - 2y - 5 =

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$(3, -1)$

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(D) The equation of the tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ is given by $x(1) + y(-2) = 5$,which simplifies to $x - 2y - 5 = 0$. This line is tangent to the circle $x^2 + y^2 - 8x + 6y + 20 = 0$. The center of this circle is $(4, -3)$. The point of contact is the foot of the perpendicular from the center $(4, -3)$ to the line $x - 2y - 5 = 0$. Let the point of contact be $(x_1, y_1)$. Using the property $\frac{x_1 - 4}{1} = \frac{y_1 + 3}{-2} = -\frac{4 - 2(-3) - 5}{1^2 + (-2)^2} = -\frac{4 + 6 - 5}{5} = -\frac{5}{5} = -1$. Thus,$x_1 - 4 = -1 \implies x_1 = 3$ and $y_1 + 3 = 2 \implies y_1 = -1$. The point of contact is $(3, -1)$.

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

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