Question

Find the equations of the circles whose centres lie on the line $4x+3y=2$ and which touch the lines $x+y+4=0$ and $7x-y+4=0$

Answer 1

Let, the equation of the circle be

$x^2+y^2+2gx+2fy+c=0$.................(1)

whose centre is $(-g,-f)$.

According to the problem $(-g,-f)$ lies on the line $4x+3y=2$, then

$-4g-3f=2$

or, $4g+3f=-2$...............(2)

Since, the lines $x+y+4=0$ and $7x-y+4=0$ touch the circle, then the distance between the centre and each lines are equal and it equals to the radius of the circle.

$\therefore$ distance between the centre of the circle (1) and the line $x+y+4=0$ is $\frac{|-g-f+4|}{\sqrt{2}}=\frac{|g+f-4|}{\sqrt{2}}$.

$\therefore$ distance between the centre of the circle (1) and the line $7x-y+4=0$ is $\frac{|-7g+f+4|}{\sqrt{50}}=\frac{|7g-f-4|}{5\sqrt{2}}$.

Also, the radius of the circle $=\sqrt{g^2+f^2-c}$.

Now, we have the following equations

$\frac{|g+f-4|}{\sqrt{2}}$ $=$ $\frac{|7g-f-4|}{5\sqrt{2}}$

$\Rightarrow 5(g+f-4)=\pm (7g-f-4)$

$\Rightarrow$ $5(g+f-4)=(7g-f-4)$ and $5(g+f-4)=-(7g-f-4)$

$\Rightarrow 2g-6f=-16$..........(3) and $3g+f=6$........(4)

Now, solving (2) and (3) we get,

$g=-2,f=2$

and solving (2) and (4) we get,

$g=4,f=-6$.

Also, $\frac{|g+f-4|}{\sqrt{2}}$ $=$ $\sqrt{g^2+f^2-c}$

or, $(g+f-4{)}^2=2(g^2+f^2-c)$..........(5)

For, $g=-2,f=2$ we get from (5),

$2(4+4-c)=(-4{)}^2$

or, $c=0$

and for $g=4,f=-6$ we get from (5),

$2(16+36-c)=(-6{)}^2$

or, $c=34$.

Then equation of the circles are

$x^2+y^2-4x+4x=0$ and $x^2+y^2+8x-12x+34=0$.

or, $(x-2{)}^2+(y+2{)}^2=8$ and $(x+4{)}^2+(y-6{)}^2=18$.