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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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Verified Answer
The correct answer is:
$x^2+y^2-4 x-4 y+7=0$
Since, the line $(x-2) \cos \theta+(y-2) \sin \theta=1$
touches a circle. So it is a tangent equation to a circle.
Equation of tangent to a circle at
$\left(x_1, y_1\right)$ is $(x-h) x_1+(y-k) y_1$
$=a^2$ to a circle
$(x-h)^2+(y-k)^2=a^2$
Then, after comparing
$\begin{aligned} & x-h=x-2 y-k=y-2 \text { and } a^2=1 \\ & x_1=\cos \theta, y_1=\sin \theta\end{aligned}$
$\therefore$ Required equation of circle
$(x-2)^2+(y-2)^2=1$
$\Rightarrow \quad x^2+y^2-4 x-4 y+7=0$
touches a circle. So it is a tangent equation to a circle.
Equation of tangent to a circle at
$\left(x_1, y_1\right)$ is $(x-h) x_1+(y-k) y_1$
$=a^2$ to a circle
$(x-h)^2+(y-k)^2=a^2$
Then, after comparing
$\begin{aligned} & x-h=x-2 y-k=y-2 \text { and } a^2=1 \\ & x_1=\cos \theta, y_1=\sin \theta\end{aligned}$
$\therefore$ Required equation of circle
$(x-2)^2+(y-2)^2=1$
$\Rightarrow \quad x^2+y^2-4 x-4 y+7=0$
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