Question

Question #2d1d2

Answer 1

1st part
Given a circle with A(5, 1) and B (8, -2) as the endpoints of the diameter. Let P(x,y) be any point on the circle. The equation of the locus of P will be the equation of circle .

AB being diameter `APB=90^@`

Hence

`"slope of PA"xx"slope of PB"=-1`

`=>(y-1)/(x-5)xx(y+2)/(x-8)=-1`

`=>(y-1)(y+2)+(x-5)(x-8)=0`

`=>y^2+y-2+x^2-13x+40=0`

`=>x^2+y^2-13x+y+38=0`

2nd part
Given a circle with (3, -4) and (6, 2) as the endpoints of the diameter.

In this case the equation of the circle will be

`(y+4)/(x-3)xx(y-2)/(x-6)=-1`

`=>(y+4)(y-2)+(x-3)(x-6)=0`

`=>y^2+2y-8+x^2-9x+18=0`

`=>x^2+y^2-9x+2y+10=0`

3rd part

given a circle with the center at the original and passing through
(4, 3).

Radius of the circle `r=sqrt(3^2+4^2)=5`

Equation of the circle

`x^2+y^2=5^2`