Question

Find the equation to the circle which touches the axis of $x$ at a distance $3$ from the origin and intercepts a distance a distance $4$ on the axis of $y$.

Answer 1

Consider the right triangle, OCB.

By Pythagoras Theorem, we have

$O{C}^2=O{B}^2+C{B}^2$

$\Rightarrow O{C}^2=4^2+3^2$

$\Rightarrow O{C}^2=16+9$

$\Rightarrow O{C}^2=25$

$\Rightarrow OC=5$

Thus, the radius of the circle is $5$ units.

The coordinates of the centre of the circle is $O(4,5)$ or $O^{\prime }(-4,5)$

Thus, the equation of the circle is

$(x-4{)}^2+(y-5{)}^2=5^2$

or $(x+4{)}^2+(y-5{)}^2=5^2$

$\Rightarrow x^2+16-8x+y^2+25-10y=25$

or $x^2+16+8x+y^2+25-10y=25$

$\Rightarrow x^2+y^2-8x-10y+16=0$

or $x^2+y^2+8x-10y+16=0$.