The equations of the tangents to the circle x | vedclass

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Question

(A) The given circle is $x^2 + y^2 = 50$ and the line is $x = -7$.Substituting $x = -7$ into the circle equation: $(-7)^2 + y^2 = 50 \implies 49 + y^2

Vedclass · Accepted Answer

$7x \pm y + 50 = 0$

Vedclass · Answer

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

The equations of the tangents to the circle $x^2 + y^2 = 50$ at the points where the line $x + 7 = 0$ meets it,are

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