The point of contact of the tangent to the ci | vedclass

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(B) 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 $xx_1 + yy_1 = r^2$.Substituting the val

Vedclass · Accepted Answer

$(3, -1)$

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(B) 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 $xx_1 + yy_1 = r^2$.Substituting the values,we get $x(1) + y(-2) = 5$,which simplifies to $x - 2y = 5$.To find the point of contact that also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$,we check which of the given options satisfies the equation of the tangent $x - 2y = 5$.For option $(B)$,$(3, -1)$: $3 - 2(-1) = 3 + 2 = 5$. This satisfies the equation.Thus,the point of contact is $(3, -1)$.

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

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