Question

What is the standard form of the equation of a circle given points: (7,-1), (11,-5), (3,-5)?

Answer 1

Standard form of circle is `(x-7)^2+(y+5)^2=16`

Explanation:

Let the equation of circle be `x^2+y^2+2gx+2fy+c=0`, whose centre is `(-g,-f)` and radius is `sqrt(g^2+f^2-c)`. As it passes though `(7,-1)`, `(11,-5)` and `(3,-5)`, we have

`49+1+14g-2f+c=0` or `14g-2f+c+50=0` ......(1)

`121+25+22g-10f+c=0` or `22g-10f+c+146=0` ...(2)

`9+25+6g-10f+c=0` or `6g-10f+c+34=0` ......(3)

Subtracting (1) from (2) we get

`8g-8f+96=0` or `g-f=-12` ......(A)

and subtracting (3) from (2) we get

`16g+112=0` i.e. `g=-7`

putting this in (A), we have `f=-7+12=5`

and putting values of `g` and `f` in (3)

`6xx(-7)-10xx5+c+34=0` i.e. `-42-50+c+34=0` i.e. `c=58`

andequation of circle is `x^2+y^2-14x+10y+58=0`

and its centre is `(7,-5)` abd radius is `sqrt(49+25-58)=sqrt16=4`

and standard form of circle is `(x-7)^2+(y+5)^2=16`

graph{x^2+y^2-14x+10y+58=0 [-3.08, 16.92, -9.6, 0.4]}