Question

A circle touches the y-axis at $ \left( 0,2 \right) $ and has an intercept of 4 units on the positive side of the x-axis. Then find the equation of the circle.
A. \[{{x}^{2}}+{{y}^{2}}-4\left( \sqrt{2}x+y \right)+4=0\]
B. \[{{x}^{2}}+{{y}^{2}}-4\left( x+\sqrt{2}y \right)+4=0\]
C. \[{{x}^{2}}+{{y}^{2}}-2\left( \sqrt{2}x+y \right)+4=0\]
D. none of these

Answer 1

Hint:
We first create the general equation of a circle. We use the point on the Y-axis to put on the equation and then use the radius value to find the distance of the centre $ \left( -g,-f \right) $ to the line $ x=0 $ and also the intercept form to find three equations. We solve those equations to find the value of $ g,f,c $ to get the equation of the circle.

Complete step by step answer:
Let’s assume the equation of the circle as $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ . The circle touches the y-axis at $ \left( 0,2 \right) $ which means the point is on the circle. We place the point in the equation and get
 $ \begin{align}
  & {{\left( 2 \right)}^{2}}+2f\left( 2 \right)+c=0 \\
 & \Rightarrow 4f+c+4=0 \\
\end{align} $
We know that for a general equation of circle $ {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 $ , the intercept on X-axis is given by $ 2\sqrt{{{g}^{2}}-c} $ which is equal to 4 as the circle has an intercept of 4 units on the positive side of the x-axis. We get
 $ \begin{align}
  & 2\sqrt{{{g}^{2}}-c}=4 \\
 & \Rightarrow {{g}^{2}}-c=4 \\
\end{align} $
The circle touches the y-axis at $ \left( 0,2 \right) $ which means the Y-axis with equation $ x=0 $ is a tangent of the circle. This means the radius of the circle is equal to the distance of the centre $ \left( -g,-f \right) $ to the line $ x=0 $ .
The formula for the general equation also gives the same statement of $ {{f}^{2}}=c $ .
Therefore, we got 3 equations for 3 unknowns which are $ 4f+c+4=0,{{g}^{2}}-c=4,{{f}^{2}}=c $ .
We solve them to get $ {{f}^{2}}+4f+4=0 $ . Solving the equation, we get $ f=-2 $ .
Placing the value, we get $ {{f}^{2}}=c=4 $ and $ {{g}^{2}}=c+4=4+4=8 $ which gives $ g=\pm 2\sqrt{2} $ .
Placing the values, we get the equation of the circles as $ {{x}^{2}}+{{y}^{2}}\pm 4\sqrt{2}x-4y+4=0 $ .
To match the given options, we get the equation as \[{{x}^{2}}+{{y}^{2}}-4\left( \pm \sqrt{2}x+y \right)+4=0\].
The correct option is A.

Note:
 We have to remember the equations and the formulas are derived form. We can directly use them to find relations. To match with given options, we eliminated the equation \[{{x}^{2}}+{{y}^{2}}-4\left( -\sqrt{2}x+y \right)+4=0\]. But that equation is also correct as a solution of the problem.