Question

How to find the standard form of the equation of the specified circle given that it Passes through points A(-1,3) y B (7,-1) and it center is on line 2x+y-11=0?

Answer 1

`(x-4)^2+(y-3)^2=5^2`

Explanation:

Let the coordinate of the center of the circle be `(a,b)" and its radius be " r`

So the equation of the circle is

`color(red)(t(x-a)^2+(y-b)^2=r^2)`
By the given condition its center lies on the straight line `2x+y-11=0`

So

`2a+b=11......(1)`

Again `A(-1,3) and B(7,-1)` are on the circle. So

`(a+1)^2+(b-3)^2= (a-7)^2+(b+1)^2`

`=>(a+1)^2+(b-3)^2-(a-7)^2-(b+1)^2=0`

`=>8(2a-6)-4(2b-2)=0`

`=>(2a-6)-(b-1)=0`

`=>2a-6-b+1=0`

`=>2a-b=5.....(2)`

Subtracting (2) from (1) we get

`2b=6=>b=3`

putting the value of b in (2)

`2a-3=5=>a=4`

Radius of the circle

`r=sqrt((a+1)^2+(b-3)^2)`

`=sqrt((4+1)^2+(3-3)^2)=5`

So equation of the circle

`(x-4)^2+(y-3)^2=5^2`