Question

Two circles pass through $(-1,4)$ and their centres lie on $x^2+y^2+2x+4y=4$. If $r_1$ and $r_2$ are maximum and minimum radii and $\frac{r_1}{r_2}=a+b\sqrt{2}$, then the value of $a+b$ is.

Answer 1

Solving for the value of $a,b$:

Step $1$: Write the equation of circle

$x^2+y^2+2x+4y=4$ is the equation of given circle

$\Rightarrow$ $x^2+2x+1+y^2+4y+4=4+1+4$

$\Rightarrow$ ${\left(x+1\right)}^2+{\left(y+2\right)}^2=3^2$

Comparing with standard centre radius form of circle ${\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2=r^2$ we get

$\left(x_1,y_1\right)=\left(-1,-2\right)$ and $r=3$

Step $2$: Transform the cartesian coordinate into polar coordinate

Any point on a given circle can be given as

$\left(x,y\right)=\left(r\cos \theta +x_{1,}r\sin \theta +y_1\right)$

Substituting the required values we get

$\left(x,y\right)=\left(3\cos \theta -1,3\sin \theta -2\right)$

The point $(-1,4)$ lies on the required circles.

Hence, the distance between $(-1,4)$ and $\left(3\cos \theta -1,3\sin \theta -2\right)$ will be equal to the radius.

Step $3$: Apply distance formula to obtain a relation between $r$ and $\theta$

$r=\sqrt{{\left(3\cos \theta -1+1\right)}^2+{\left(3\sin \theta -2-4\right)}^2}$

$\Rightarrow$ $r=\sqrt{\left(9{\cos }^2\theta \right)+\left(9{\sin }^2\theta +36–36\sin \theta \right)}$

$\Rightarrow$ $r=\sqrt{45-36\sin \theta }$ $...\left[{\sin }^{2}x+{\cos }^{2}x=1\right]$

$r$ will be maximum when $\sin \theta =-1$ and $r$ will be minimum when $\sin \theta =1$

$\Rightarrow$ $r_1=\sqrt{81}=9$ and $r_2=\sqrt{9}=3$

$\Rightarrow$ $\frac{r_1}{r_2}=3=a+b\sqrt{2}$

$\Rightarrow$ $a=3,b=0$

Hence, the value of $a+b$ is $3$.