If the centre of a circle is (-6, 8) and it p | vedclass

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(B) The centre of the circle is $C(-6, 8)$.Since the circle passes through the origin $O(0, 0)$,the radius $OC$ is the line segment connecting $(0, 0)

Vedclass · Accepted Answer

$3x - 4y = 0$

Vedclass · Answer

(B) The centre of the circle is $C(-6, 8)$.Since the circle passes through the origin $O(0, 0)$,the radius $OC$ is the line segment connecting $(0, 0)$ and $(-6, 8)$.The slope of the radius $OC$ is $m_{radius} = \frac{8 - 0}{-6 - 0} = \frac{8}{-6} = -\frac{4}{3}$.The tangent at the origin is perpendicular to the radius $OC$.Therefore,the slope of the tangent $m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{-4/3} = \frac{3}{4}$.The equation of the tangent passing through the origin $(0, 0)$ with slope $m = \frac{3}{4}$ is $y - 0 = \frac{3}{4}(x - 0)$,which simplifies to $4y = 3x$ or $3x - 4y = 0$.

If the centre of a circle is $(-6, 8)$ and it passes through the origin,then the equation of its tangent at the origin is

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