The equation to two circles which touch the $Y$-axis at $(0,3)$ and make an intercept of 8 units on $X$-axis are

• A

  $x^{2}+y^{2} \pm 6 x-10 y+9=0$
• Correct answer
  B

  $x^{2}+y^{2} \pm 10 x-6 y+9=0$
• C

  $x^{2}+y^{2}+10 x \pm 6 y+9=0$
• D

  $x^{2}+y^{2}-8 x \pm 10 y+9=0$

Solution

Equation of circle will be $$ (x \pm a)^{2}+(y-3)^{2}=a^{2} $$ Now, in $\triangle A C D$, $$ \begin{aligned} & A C^{2}=A D^{2}+C D^{2} \\ \Rightarrow \quad a^{2}=(4)^{2}+(3)^{2} \\ & {\left[\therefore A C=a, A D=\frac{1}{2} A B=4, C D=3\right] } \\ \Rightarrow \quad & a^{2}=25 \Rightarrow a=5 \end{aligned} $$ So, equation of circles will be $$ \begin{aligned} (x \pm 5)^{2}+(y-3)^{2} &=5^{2} \\ \Rightarrow x^{2} \pm 10 x+25+y^{2}-6 y+9 &=25 \\ \Rightarrow \quad x^{2}+y^{2} \pm 10 x-6 y+9 &=0 \end{aligned} $$

Hence, the correct answer is option: 'B'.