The equations of tangents to the circle x^2+y | vedclass

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(C) Let the equation of the line passing through $(4, -2)$ be $y - (-2) = m(x - 4)$,which simplifies to $mx - y - (4m + 2) = 0$.For this line to be a

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$3x+y=10, x-3y=10$

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(C) Let the equation of the line passing through $(4, -2)$ be $y - (-2) = m(x - 4)$,which simplifies to $mx - y - (4m + 2) = 0$.For this line to be a tangent to the circle $x^2 + y^2 = 10$,the perpendicular distance from the center $(0, 0)$ to the line must be equal to the radius $r = \sqrt{10}$.Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\sqrt{10} = \frac{|m(0) - 1(0) - (4m + 2)|}{\sqrt{m^2 + (-1)^2}}$.$\sqrt{10} = \frac{|4m + 2|}{\sqrt{m^2 + 1}}$.Squaring both sides: $10(m^2 + 1) = (4m + 2)^2$.$10m^2 + 10 = 16m^2 + 16m + 4$.$6m^2 + 16m - 6 = 0$,which simplifies to $3m^2 + 8m - 3 = 0$.$(3m - 1)(m + 3) = 0$,so $m = \frac{1}{3}$ or $m = -3$.For $m = \frac{1}{3}$,the line is $y + 2 = \frac{1}{3}(x - 4) \implies 3y + 6 = x - 4 \implies x - 3y = 10$.For $m = -3$,the line is $y + 2 = -3(x - 4) \implies y + 2 = -3x + 12 \implies 3x + y = 10$.Thus,the equations are $3x + y = 10$ and $x - 3y = 10$.

The equations of tangents to the circle $x^2+y^2=10$ from the point $(4,-2)$ are

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