A circle is such that (x-2) cos theta + (y-2) | vedclass

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(A) The given equation of the line is $(x-2) \cos 	heta + (y-2) \sin 	heta = 1$. This line represents a tangent to the circle for all values of $	h

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$x^2+y^2-4x-4y+7=0$

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(A) The given equation of the line is $(x-2) \cos 	heta + (y-2) \sin 	heta = 1$. This line represents a tangent to the circle for all values of $	heta$. The distance from the center $(h, k)$ of the circle to the tangent line must be equal to the radius $r$. Rewriting the line as $(x-2) \cos 	heta + (y-2) \sin 	heta - 1 = 0$,the distance from $(h, k)$ is $\frac{|(h-2) \cos 	heta + (k-2) \sin 	heta - 1|}{\sqrt{\cos^2 	heta + \sin^2 	heta}} = r$. Since this holds for all $	heta$,we must have $h-2 = 0$ and $k-2 = 0$,which gives the center $(2, 2)$. Then,the distance becomes $|-1| = r$,so $r = 1$. The equation of the circle is $(x-2)^2 + (y-2)^2 = 1^2$. Expanding this,we get $x^2 - 4x + 4 + y^2 - 4y + 4 = 1$. Thus,$x^2 + y^2 - 4x - 4y + 7 = 0$.

$A$ circle is such that $(x-2) \cos \theta + (y-2) \sin \theta = 1$ touches it for all values of $\theta$. Then,the circle is

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