$x^2+y^2-2(t^2-3t+1)x-2(t^2+2t)y+t=0$,where $t$ is a parameter.If the power of a point $P(a,b)$ wrt the circle $C$ is constant

Question

A variable circle $C$ has the equation $x^2+y^2-2(t^2-3t+1)x-2(t^2+2t)y+t=0$,where $t$ is a parameter.If the power of a point $P(a,b)$ wrt the circle $C$ is constant then the ordered pair $(a,b)$ is

$(A)(\frac{1}{10},\frac{-1}{10})\hspace{1cm}(B)(\frac{-1}{10},\frac{1}{10})\hspace{1cm}(C)(\frac{1}{10},\frac{1}{10})\hspace{1cm}(D)(\frac{-1}{10},\frac{-1}{10})$

If the options are given in this question,i can solve this question easily,and find out the answer among the options.

My question/doubt is that can this question still be solved if the options are not given?.Is there a proper method to solve it without options given.

Answer 1

The power of $P(a, b)$ wrt the circle is constant means

$a^2+b^2-2(t^2-3t+1)a-2(t^2+2t)b+t= constant$.

The equation can be re-written as $()t^2 + []t = constant$

Setting $() = 0$ gives $a = - b$

Setting $[] = 0$ gives $6a – 4b + 1 = 0$

Result follows.