Question

The equation of the circle which touches $x-$axis at $(3,0)$ and passes through $(1,4)$ is given by

• A. $x^2+y^2-6x-5y+9=0$
• B. $x^2+y^2+6x+5y-9=0$
• C. $x^2+y^2-6x-5y-9=0$
• D. $x^2+y^2+6x-5y+9=0$

Answer 1

The correct option is A $x^2+y^2-6x-5y+9=0$

Let the eq of circle be

$x^2+y^2+2gx+2\beta y+c=0$

now, x intercept = $2\sqrt{g^2-c}$

But circle touches x axis $\therefore$ x intercept = 0

$\therefore 2\sqrt{g^2-c}=0\therefore g^2=c$

$\therefore e{q}^n$ become $x^2+y^2+3gx+2\beta y+g^2=0$

Now circle passes through (3, 0) and (1, 4)

$\therefore (3{)}^2+(0{)}^2+2g\left(3\right)+0+g^2=0$

$9+6g+g^2=0$

$\therefore g=-3$

Also $(1{)}^2+(4{)}^2+2(-3)\left(1\right)+2\beta \left(4\right)+9=0$

$\therefore 1+16-6+8\beta +9=0$

$\therefore 20+8\beta =0$ $\therefore \beta =\frac{5}{2}$

Hence $e{q}^n$ of circle is

$x^2+y^2-6x-5y+9=0$