Question

Find the equation of a circle which touches the positive axis of $y$ at a distance $3$ from the origin and intercepts a distance $6$ on the axis of $x.$

Answer 1

the circle touches $y$ axis at $3$ from origin.

$\Rightarrow (0,3)$ is a point on the circle.

and intercept at a point on the circle.

and intercepts at a distance $6$ on x-axis.

$\Rightarrow (6,0)$ is also a point.

$A=\left(0.3\right)$ $B(6,0)$

$0=(p,v)$

$AC$ is tangent,

$\Rightarrow AO$ is perpendicular to $AC$

and $AO$ is parallel to x-axis. (i.e $y$ coordinate doesn't change)

$\Rightarrow V=3$

$OA=OB=radius$

$\Rightarrow \sqrt{(p-o{)}^2+(3-3{)}^2}=\sqrt{(p-6{)}^2+(3-0{)}^2}$

$\Rightarrow p^2=p^2+36-12p+q$

$\Rightarrow 12p=45$

$\Rightarrow P=\frac{45}{4}$

radius =$P\frac{15}{4}$ and centre $=(\frac{15}{4},3)$

$\therefore$ equation of cicle is ,

$(x-\frac{15}{4}{)}^2+(y-3{)}^2=(\frac{15}{4}{)}^2$

$\Rightarrow x^2-\frac{15x}{2}+y^2+9-6xy=0$

$\Rightarrow 2x^2+2y^2-15x-12y+18=0$