The tangent to the circle x^2+y^2=5 at (1,-2) | vedclass

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Question

(A) The equation of the tangent to the circle $x^2+y^2=5$ at the point $(x_1, y_1) = (1, -2)$ is given by $x x_1 + y y_1 = r^2$. Substituting the valu

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$(3,-1)$

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(A) The equation of the tangent to the circle $x^2+y^2=5$ at the point $(x_1, y_1) = (1, -2)$ is given by $x x_1 + y y_1 = r^2$. Substituting the values,we get $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$. For the second circle $x^2 + y^2 - 8x + 6y + 20 = 0$,the center is $(4, -3)$ and the radius is $\sqrt{4^2 + (-3)^2 - 20} = \sqrt{16 + 9 - 20} = \sqrt{5}$. The point of contact on the second circle must lie on the line $x - 2y = 5$. Checking the options: For $(3, -1)$: $3 - 2(-1) = 3 + 2 = 5$. This point satisfies the tangent equation. For $(3, 1)$: $3 - 2(1) = 1 
eq 5$. For $(-3, -1)$: $-3 - 2(-1) = -1 
eq 5$. For $(-3, 1)$: $-3 - 2(1) = -5 
eq 5$. Thus,the point of contact is $(3, -1)$.

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

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