Question

Find the equation of the circle :
(a) centered at $(3,-2)$ with radius $4$
(b) with end points of the diameter as $(2,-1)$ and $(3,2)$
(c) with parametric co-ordinates $x=-3+4\cos \theta ,y=4+4\sin \theta$
(d) passing through three points $(0,2),(3,0)$ and $(3,2)$

Answer 1

$1.(x-3{)}^2+(y+2{)}^2=16$

$2.$ Centre of the circle

$x_{centre}=\frac{3+2}{2}=\frac{5}{2}$

$y_{centre}=\frac{-1+2}{2}=\frac{1}{2}$

Using distance formula to get radius

$r=\sqrt{(2-2.5{)}^2+(-1-0.5{)}^2}$

$r=\frac{\sqrt{10}}{2}$

Equation of circle = $(x-2.5{)}^2+(y-0.5{)}^2=\frac{\sqrt{10}}{2}$

$3.(x+3)=4cos\theta ,(y-4)=4sin\theta$

Squaring and adding

$(x+3{)}^2+(y-4{)}^2=16$

$4.$Let the general equation be

$x^2+y^2+Dx+Ey+F=0$

Putting $(0,2)$

$4+2E+F=0$......(i)

Putting $(3,0)$

$9+3D+F=0$.......(ii)

Putting $(3,2)$

$13+3D+2E+F=0$.....(iii)

From (i) & (ii)

$13+3D+2E+2F=0$

or $F=0$ from (iii)

Hence $E=-2$

$F=-3$