The equations of the tangents to the circle x | vedclass

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Question

(A) Given the circle equation $x^2 + y^2 = 13$. Let the point on the circle be $(2, y')$. Substituting $x = 2$ into the circle equation: $2^2 + (y')^2

Vedclass · Accepted Answer

$2x + 3y = 13, 2x - 3y = 13$

Vedclass · Answer

(A) Given the circle equation $x^2 + y^2 = 13$. Let the point on the circle be $(2, y')$. Substituting $x = 2$ into the circle equation: $2^2 + (y')^2 = 13$ $4 + (y')^2 = 13$ $(y')^2 = 9$ $y' = \pm 3$. So,the points are $(2, 3)$ and $(2, -3)$. 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 $(2, 3)$: $2x + 3y = 13$. For point $(2, -3)$: $2x - 3y = 13$. Thus,the equations of the tangents are $2x + 3y = 13$ and $2x - 3y = 13$.

The equations of the tangents to the circle $x^2 + y^2 = 13$ at the points whose abscissa is $2$,are

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