Question

Equation of circle which touch x-axis at $(3,0)$ and making y-intercept of length $8$ units is?

• A. $x^2+y^2-6x\pm 10y+9=0$
• B. $x^2+y^2-6x\pm 10y-9=0$
• C. $x^2+y^2\pm 6x\pm 10y+9=0$
• D. None

Answer 1

The correct option is A $x^2+y^2-6x\pm 10y+9=0$

$\begin{array}{c}{a}^2=3^2+4^2\\ a=\sqrt{25}=5\end{array}$

one centre at $(5,3)$ and other at $(-5,3)$

So, one radius is 5 and other is -5

$\begin{array}{c}\therefore Equationoflineis\\ \left(i\right){\left(x-3\right)}^2+{\left(y-5\right)}^2=5^2\\ x^2+9-6x+y^2+25-10y=25\\ x^2+y^2-6x-10y+9=0\\ \left(ii\right){\left(x-3\right)}^2+{\left(y+5\right)}^2=5^2\\ x^2+9-6x+y^2+25+10y=25\\ x^2+y^2-6x+10y+9=0\end{array}Z$

Hence, equation is

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