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

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

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(A) The equation of the tangent to the circle $x^2 + y^2 = 5$ at $(1, -2)$ is given by $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$,or $x = 2y + 5$.Substitute $x = 2y + 5$ into the equation of the second circle $x^2 + y^2 - 8x + 6y + 20 = 0$:$(2y + 5)^2 + y^2 - 8(2y + 5) + 6y + 20 = 0$Expanding the terms:$(4y^2 + 20y + 25) + y^2 - 16y - 40 + 6y + 20 = 0$Combining like terms:$5y^2 + 10y + 5 = 0$Dividing by $5$:$y^2 + 2y + 1 = 0$$(y + 1)^2 = 0$This gives $y = -1$. Substituting $y = -1$ into $x = 2y + 5$ gives $x = 2(-1) + 5 = 3$.Since there is only one point of intersection $(3, -1)$,the line touches the circle.

List-$I$	List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$	$I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$	$II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$	$III.$ $60^{\circ}$
	$IV.$ $30^{\circ}$

The tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ intersects the circle $x^2 + y^2 - 8x + 6y + 20 = 0$ at which of the following?

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