Question

The equation of the circle which touches x-axis at $(0,0)$ and touches the line $3x+4y-5=0$ is

• A. $x^2+y^2-4y=0$
• B. $x^2+y^2-10y=0$
• C. $x^2+y^2+10x=0$
• D. $x^2+y^2+10y=0$

Answer 1

The correct option is D $x^2+y^2+10y=0$

Equation of circle touching x-axis at $(0,0),$ means centre of circle lie on Y-axis i.e. $(0,k)$.

$(x-0{)}^2+(y-k{)}^2=k^2$

$S:x^2+y^2-2ky=0......\left(1\right)$

Circle S touches line $3x+4y-5=0$
Perpendicular distance from the centre (0,k) of circle to the given line is equal to the radius of the circle

$\therefore k=\left|\frac{3\left(0\right)+4k-5}{\sqrt{3^2+4^2}}\right|=\left|\frac{4k-5}{5}\right|$

$5k=4k-5$

$k=-5$

$\therefore$ Equation of circle is

$x^2+y^2-(-5)\times 2y=0$

$\Rightarrow x^2+y^2+10y=0$