Question

The circle S has the equation${{(x-4)}^{2}}+{{(y-2)}^{2}}=13.$ The point (P, 0) lies on S. Find the two real values of P.

Answer 1

Hint: Any point lying on the circle satisfies the given equation of the circle. So put the given point in the equation of the circle and solve it. Using this concept we will get the correct answer.

Complete step by step solution:
The given equation of circle
$S:{{(x-4)}^{2}}+{{(y-2)}^{2}}=13-(1)$
As point (P,0) lies on circle S so (P,0) satisfy the equation (1) hence we substitute $x=P,\text{ y=0}$ in the equation (1) so we can write
${{(P-4)}^{2}}+{{(0-2)}^{2}}=13$
Now expanding using formula
${{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
 we can write
\[\begin{align}
  & \Rightarrow {{P}^{2}}+{{(-4)}^{2}}+2P(-4)+{{(-2)}^{2}}=13 \\
 & \Rightarrow {{P}^{2}}+16-8P+4-13=0 \\
 & \Rightarrow {{P}^{2}}-8P+7=0 \\
\end{align}\]
Here we get a quadratic equation so we have to find here the roots of quadratic equation
As we know
$D={{b}^{2}}-4ac$
Here $a=1,b=-8,c=7$
 so
$\begin{align}
  & D={{(-8)}^{2}}-4(1)(7) \\
 & \Rightarrow D=64-28 \\
 & \Rightarrow D=36 \\
\end{align}$
Since
$D\ge 0$
Hence roots are real.
Now factorise this quadratic equation by splitting middle term and we can write
$\begin{align}
  & \Rightarrow {{P}^{2}}-8P+7=0 \\
 & \Rightarrow {{P}^{2}}-7P-P+7=0 \\
 & \Rightarrow P(P-7)-1(P-7)=0 \\
 & \Rightarrow (P-1)(P-7)=0 \\
 & \Rightarrow P=1 \\
 & or\text{ P=7} \\
\end{align}$
So we have two value of P
$\begin{align}
  & P=1 \\
 & P=7 \\
\end{align}$

Note: To solve this type of problem students should know that we can find the roots of quadratic equations by using the following formula and this is an alternative method for this problem.
$\Rightarrow {{x}_{1,2}}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$