Find the equation of the circle touching the line $(x-2)\cos\theta+(y-2)\sin\theta=1$ for all values of $\theta$

Question

Find the equation of the circle touching the line $(x-2)\cos\theta+(y-2)\sin\theta=1$ for all values of $\theta$

The answer is given on toppr website.

It says, $(x-2)\cos\theta+(y-2)\sin\theta=\cos^2\theta+\sin^2\theta$, then comparing coefficients, it says,

$x-2=\cos\theta, y-2=\sin\theta$

I am not sure about this step. Do you think this step is valid?

Another answer exists on sarthaks website.

It says equation of tangent to the circle $(x-h)^2+(y-k)^2=a^2$ at $(x_1,y_2)$ is

$(x-h)x_1+(y-k)y_1=a^2$

Is this valid? I am not sure about this.

Answer 1

Here's a simpler and correct (verified by GeoGebra) solution.

The circle center is $(m,n)$, then its distance to the line must be constant as $\theta$ varies. If we write the line's equation as $$x\cos\theta+y\sin\theta-2\cos\theta-2\sin\theta-1=0,$$ apply the distance formula to have $$d=\frac{|m\cos\theta+n\sin\theta-2\cos\theta-2\sin\theta-1|}{\sqrt{\cos^2\theta+\sin^2\theta}}.$$ The denominator is $\sqrt1=1$, and the numerator is can be written as $$\left|\sqrt{(m-2)^2+(n-2)^2}\sin(\theta+\varphi)-1\right|.$$ For it to be constant, $\sqrt{(m-2)^2+(n-2)^2}=0$, so $m=n=2$, and so $d=1$.

Our circle is therefore $\odot((2,2),1)$, equation $$(x-2)^2+(y-2)^2=1.$$

Answer 2

I thought of another solution. Let us find the family of tangent lines of the circle $$C: (x-m)^2+(y-n)^2=r^2.\tag1$$ Let $P=(m+r\cos\theta,n+r\sin\theta)\in C$ where $\theta\in[0,2\pi)$ is the family parameter. Then the slope of the tangent line at $P$ is $y'=\left.-\frac{x-m}{y-n}\right\rvert_P=-\cot\theta$. Hence, the tangent line at point P is $$(x-m)\cos\theta+(y-n)\sin\theta=r.\tag2$$ Check! So, the envelope of the family $(2)$ is the circle $(1)$.

Therefore, the envelope of the family $(x-1)\cos\theta+(y-1)\sin\theta=1$ is $(x-1)^2+(y-1)^2=1$.

Answer 3

You can see that the family of lines is always at a distance $1$ from $(2,2)$.

Hence this family is tangent to $(x-2)^2+(y-2)^2=1$