Question

The equation of two circles which touch the $y$ - axis at $\left(0,3\right)$ and make an intercept of $8$ units on $x$ - axis , are

• A. $x^2+y^2\pm 10x-6y+9=0$
• B. $x^2+y^2\pm 6x-10y+9=0$
• C. $x^2+y^2-8x\pm 10y+9=0$
• D. $x^2+y^2-10x\pm 6y+9=0$

Answer 1

The correct option is A

$x^2+y^2\pm 10x-6y+9=0$

Given Data:

Equation of two circles touch the $y$ - axis at $\left(0,3\right)$

Intercept = $8$ units on $x$ - axis

Find the equation of the circle:

In the given figure, the circle touches the $y$-axis at $\left(0,3\right)$.

From triangle $OAB$, $r$ is the radius of circle.

$\begin{array}{rcl}r& =& \sqrt{3^2+4^2}\\ & =& \sqrt{25}\\ & =& 5\end{array}$

Explanation for Correct Option:

Option (A):

From option $\left(A\right)$, $x^2+y^2\pm 10x-6y+9=0$.

$\begin{array}{rcl}{0}^2+3^2\pm 10\left(0\right)-6\left(3\right)+9& =& 0\\ 0+9\pm 0-18+9& =& 0\\ 0& =& 0\end{array}$

Explanation for Incorrect Option:

Option (B):

$$\begin{gathered} x^2+y^2\pm 6x-10y+9=0 \\ {0}^2+3^2\pm 6\left(0\right)-10\left(3\right)+9=0 \\ 9-30+9=0 \\ -12\ne 0 \end{gathered}$$

Option (C):

$$\begin{gathered} x^2+y^2-8x\pm 10y+9=0 \\ {0}^2+3^2-8\left(0\right)\pm 10\left(3\right)+9=0 \\ 9\pm 30+9=0 \\ 48or-12\ne 0 \end{gathered}$$

Option (D):

$$\begin{gathered} x^2+y^2-10x\pm 6y+9=0 \\ {0}^2+3^2-10\left(0\right)\pm 6\left(3\right)+9=0 \\ 9\pm 18+9=0 \\ 36\ne 0 \end{gathered}$$

Hence, $x^2+y^2\pm 10x-6y+9=0$ are the two required equations and therefore option $\left(A\right)$ is correct.