Find the equations of the tangents drawn to t | vedclass

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Question

(A) Given the circle equation $x^2+y^2=50$ and the line $x+7=0$. Substituting $x = -7$ into the circle equation: $(-7)^2 + y^2 = 50$ $49 + y^2 = 50$ $

Vedclass · Accepted Answer

$7x+y+50=0 	ext{ and } 7x-y+50=0$

Vedclass · Answer

(A) Given the circle equation $x^2+y^2=50$ and the line $x+7=0$. Substituting $x = -7$ into the circle equation: $(-7)^2 + y^2 = 50$ $49 + y^2 = 50$ $y^2 = 1 \Rightarrow y = \pm 1$. Thus,the points of intersection are $P_1(-7, 1)$ and $P_2(-7, -1)$. The equation of the tangent to the circle $x^2+y^2=r^2$ at $(x_1, y_1)$ is given by $xx_1 + yy_1 = r^2$. For point $(-7, 1)$: $-7x + y = 50 \Rightarrow 7x - y + 50 = 0$. For point $(-7, -1)$: $-7x - y = 50 \Rightarrow 7x + y + 50 = 0$. Therefore,the required equations are $7x+y+50=0$ and $7x-y+50=0$.

Find the equations of the tangents drawn to the circle $x^2+y^2=50$ at the points where the line $x+7=0$ meets it.

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