Question

How do you find the equation for a circle touching y – axis and passing through the points (1, 5), (8, 12)?

Answer 1

Hint: Now we have the equation of the circle is ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$ . Now we know that the circle touches the y axis hence we have the radius of the circle is ${{x}_{1}}$ . Now we will substitute $r={{x}_{1}}$ and form the equation of the circle. Now we know that the points (1, 5) and (8, 12) lie on the circle. Hence we will substitute these points in the equation of the circle and hence we will get two simultaneous equations. We will solve this equation to find ${{x}_{1}},{{y}_{1}}$ and hence substitute it in the equation of the circle.

Complete step-by-step answer:
Now we know that the general equation of the circle is given ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$ .
Where r is the radius of the circle and $\left( {{x}_{1}},{{y}_{1}} \right)$ is the center of the circle.
Now note that if the circle touches y axis then the radius of the circle is nothing but ${{x}_{1}}$ as ${{x}_{1}}$ is the distance between y axis and center.
Now substituting the value of r in the general equation of circle we get,
$\Rightarrow {{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{x}_{1}}^{2}$
Now we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ hence using this we get the equation of circle as,
$\begin{align}
  & \Rightarrow {{x}^{2}}-2{{x}_{1}}x+{{x}_{1}}^{2}+{{y}^{2}}-2{{y}_{1}}y+{{y}_{1}}^{2}={{x}_{1}}^{2} \\
 & \Rightarrow {{x}^{2}}-2{{x}_{1}}x+{{y}^{2}}-2{{y}_{1}}y+{{y}_{1}}^{2}=0.................\left( 1 \right) \\
\end{align}$
Now we know that (1, 5) lies on the circle. Hence substituting x = 1 and y = 5 in equation (1) we get,
$\begin{align}
  & \Rightarrow 1-2{{x}_{1}}+25-10{{y}_{1}}+{{y}_{1}}^{2}=0 \\
 & \Rightarrow -2{{x}_{1}}-10{{y}_{1}}+{{y}_{1}}^{2}+26=0........\left( 2 \right) \\
\end{align}$
Similarly we have (8, 12) lies on the equation. Hence substituting it in equation (1) we get,
$\begin{align}
  & \Rightarrow 64-16{{x}_{1}}+144-24{{y}_{1}}+{{y}_{1}}^{2}=0 \\
 & \Rightarrow -16{{x}_{1}}-24{{y}_{1}}+{{y}_{1}}^{2}+208=0........\left( 3 \right) \\
\end{align}$
Now multiplying equation (2) by – 8 and then adding with equation (3) we get,
$\begin{align}
  & \Rightarrow 16{{x}_{1}}+80{{y}_{1}}-8{{y}_{1}}^{2}-208-16{{x}_{1}}-24{{y}_{1}}+{{y}_{1}}^{2}+208=0 \\
 & \Rightarrow 56{{y}_{1}}=7{{y}_{1}}^{2} \\
\end{align}$
Hence we get ${{y}_{1}}=8$
Now substituting ${{y}_{1}}=8$ in equation (2) we get, $-2{{x}_{1}}-80+64+26=0\Rightarrow {{x}_{1}}=5$
Now again substituting the values ${{x}_{1}}=5$ and ${{y}_{1}}=8$ in equation ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{x}_{1}}^{2}$ we get, the equation of circle as ${{\left( x-5 \right)}^{2}}+{{\left( y-8 \right)}^{2}}=25$
Hence we have the equation of required circle as ${{\left( x-5 \right)}^{2}}+{{\left( y-8 \right)}^{2}}=25$

Note: Note that when the equation $56{{y}_{1}}=7{{y}_{1}}^{2}$ is quadratic equation in ${{y}_{1}}$ which has two roots. We can see that one root of the equation is 0. But this cannot be as this would mean the center of the circle lies on y axis and we have the circle also touches y axis and hence this case is not possible.