Question

Find the power of the point $P\left( -1,1 \right)$ with respect to the circle ${{x}^{2}}+{{y}^{2}}-6x+4y-12=0$.

Answer 1

Hint: To find the power of the given point with respect to the given circle we need to substitute the value of the point given in the equation of the circle. We will put $x=-1$ and $y=1$ in the given equation of circle. Then simplifying the obtained equation we will get the desired answer.

Complete step by step solution:
We have been given an equation of circle ${{x}^{2}}+{{y}^{2}}-6x+4y-12=0$.
We have to find the power of the point $P\left( -1,1 \right)$ with respect to the circle.
Now, we know that the power of a point is the real number which represents the relative distance of a point from the circle.
To find the power of any point with respect to the circle we have to put the given coordinates in to the equation of the circle.
Here we have a point $P\left( -1,1 \right)$, so we will substitute the value $x=-1$ and $y=1$ in the given equation of circle. Then we will get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+{{y}^{2}}-6x+4y-12=0 \\
 & \Rightarrow {{\left( -1 \right)}^{2}}+{{1}^{2}}-6\left( -1 \right)+4\times 1-12=0 \\
\end{align}\]
Now, simplifying the above obtained equation we will get
\[\begin{align}
  & \Rightarrow 1+1+6+4-12=0 \\
 & \Rightarrow 12-12=0 \\
\end{align}\]
Hence we get the power of point $P\left( -1,1 \right)$ with respect to the circle ${{x}^{2}}+{{y}^{2}}-6x+4y-12=0$ as 0.

Note: The point to be remembered is that we can also find the position of point by using the power of point. If the value of power of point is negative then the point lies inside the circle, if the value of power of point is positive then the point lies outside the circle and if the value of power of point is zero then the point lies on the circle.