Question

Find the equation to a circle of radius $r$ which touches the axis of $y$ at a point distant $h$ from the origin, the centre of the circle being in the positive quadrant.
rove also that the equation to the other tangent which passes through the origin is :
$(r^{2} − h^{2}) x + 2rhy = 0$.

Answer 1

From the figure, equation of circle will be

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

Point $(0,h)$ lies on the circle so,

$k^2=r^2$

$\therefore (x-r{)}^2+(y-h{)}^2=r^2$ is the equation of circle

Let $y=mx+c$ be the other tangent

Now, it passes through $(0,0)$

$\therefore c=0$

For a line to be tangent to a circle, the distance of the centre from the line will be equal to the radius

$\frac{h-mr}{\sqrt{1+m^2}}=r$

$m=\frac{h^2-r^2}{2hr}$

So, the equation of tangent is

$x(r^2-h^2)+2hry=0$

hence proved