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  • A variable circle has the equation <img alt="\mathrm{x^{2}+y^{2}-2\left(t^{2}-3 t+1\right) x-2

A variable circle \mathrm{C} has the equation
\mathrm{x^{2}+y^{2}-2\left(t^{2}-3 t+1\right) x-2\left(t^{2}+2 t\right) y+t=0, \text { where } t \text { is a parameter. }}
If the power of point \mathrm{\mathrm{P}(\mathrm{a}, \mathrm{b})} w.r.t. the circle \mathrm{C} is constant then the ordered pair \mathrm{(a, b)} is

Option: 1

\left(\frac{1}{10},-\frac{1}{10}\right)


Option: 2

\left(-\frac{1}{10}, \frac{1}{10}\right)


Option: 3

\left(\frac{1}{10}, \frac{1}{10}\right)


Option: 4

\left(-\frac{1}{10},-\frac{1}{10}\right)


Answers (1)

best_answer

Power: \mathrm{P}=\mathrm{a}^{2}+\mathrm{b}^{2}-2\left(\mathrm{t}^{2}-3 \mathrm{t}+1\right) \mathrm{a}-2\left(\mathrm{t}^{2}+2 \mathrm{t}\right) \mathrm{b}+\mathrm{t}
                  \mathrm{=-(2 a+2 b) t^{2}+(6 a-4 b+1) t+a^{2}+b^{2}-2 a}

This power is independent of the parameter \mathrm{t}

if and only if \mathrm{2 \mathrm{a}+2 \mathrm{~b}=0 \quad \Rightarrow \quad \mathrm{a}=-\mathrm{b}}

\mathrm{\text { and } \quad 6 a-4 b+1=0 }
\mathrm{\Rightarrow \quad a=-\frac{1}{10} \text { and } b=\frac{1}{10} }

Posted by

Deependra Verma

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