The tangent to the circle x^2 + y^2 = 10 at t | vedclass

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(D) $1$. The equation of the tangent to the circle $x^2 + y^2 = r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. For $x^2 + y^2 = 10$ at $(3, 1)$, t

Vedclass · Accepted Answer

$101$

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(D) $1$. The equation of the tangent to the circle $x^2 + y^2 = r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$. For $x^2 + y^2 = 10$ at $(3, 1)$, the tangent is $3x + y = 10$, or $3x + y - 10 = 0$.$2$. The second circle is $x^2 + y^2 - 2\sqrt{10}x - 20y + k = 0$. Its center is $(\sqrt{10}, 10)$ and radius $R = \sqrt{(\sqrt{10})^2 + 10^2 - k} = \sqrt{110 - k}$.$3$. The distance from the center $(\sqrt{10}, 10)$ to the line $3x + y - 10 = 0$ must equal the radius $R$.$4$. Distance $d = \frac{|3(\sqrt{10}) + 1(10) - 10|}{\sqrt{3^2 + 1^2}} = \frac{3\sqrt{10}}{\sqrt{10}} = 3$.$5$. Setting $d^2 = R^2$, we get $3^2 = 110 - k$, so $9 = 110 - k$, which gives $k = 101$.

The tangent to the circle $x^2 + y^2 = 10$ at the point $(3, 1)$ touches the circle $x^2 + y^2 - 2\sqrt{10}x - 20y + k = 0$. Find the value of $k$.

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