Question

Find the equation of the circle that has the same centre as the circle $x^2+y^2=1$ has and passes through the point $(4,3)$.

Answer 1

The centre of the circle $x^2+y^2=1$ is $(0,0)$.

Now the equation of the circle which has the centre is $(0,0)$ is $x^2+y^2=a^2$. Where $a$ is a constant.

According to the problem this circle passes through the point $(3,4)$.

Then,

$3^2+4^2=a^2$

or, $25=a^2$

So, the equation of the required circle is $x^2+y^2=5^2$.