Question

Equation of circle passing through (-1,-2) and concentric with the circle x^2 +y^2 +3x + 4y +1 =0 ?

Answer 1

See process below

Explanation:

A circle centered in `(a,b)` with radius `r` has the equation

`(x-a)^2+(y-b)^2=r^2`

Lets complete the squares in given circle

`x^2+y^2+3x+4y+1=(x+3/2)^2+(y+2)^2-4-9/4+1=`

`=(x+3/2)^2+(y+2)^2-21/4=0` and then

`=(x+3/2)^2+(y+2)^2=21/4`

The given circle has center in `(-3/2,-2)` and has radius `sqrt21/2`

Our circle equation will be `(x+3/2)^2+(y+2)^2=r^2`

But is passing through `(-1,-2)` then

`(-1+3/2)^2+(-2+2)^2=r^2`

`1/2^2=r^2`

The requested equation is `(x+3/2)^2+(y+2)^2=1/4`