The line (x - a)cos alpha + (y - b)sin alpha | vedclass

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(C) The equation of the circle is $(x - a)^2 + (y - b)^2 = r^2$,which has center $(a, b)$ and radius $r$.The perpendicular distance $d$ from the cente

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For all values of $\alpha$

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(C) The equation of the circle is $(x - a)^2 + (y - b)^2 = r^2$,which has center $(a, b)$ and radius $r$.The perpendicular distance $d$ from the center $(a, b)$ to the line $(x - a)\cos \alpha + (y - b)\sin \alpha - r = 0$ is given by:$d = \frac{|(a - a)\cos \alpha + (b - b)\sin \alpha - r|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}}$$d = \frac{|0 + 0 - r|}{\sqrt{1}} = |-r| = r$.Since the perpendicular distance from the center to the line is equal to the radius $(d = r)$,the line is a tangent to the circle for all values of $\alpha$.

The line $(x - a)\cos \alpha + (y - b)\sin \alpha = r$ will be a tangent to the circle $(x - a)^2 + (y - b)^2 = r^2$:

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