Question

Find the equation of circle passing through a point (1,1) (2,-1) and (3,2)

Answer 1

Answer:

The equation of the circle is $(x-\frac{5}{2})^2+(y-\frac{1}{2})^2=\frac{5}{2}$.

Step-by-step explanation:

The general form of a circle is

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is center of the circle and r is radius.

It is given that circle passing through the points (1,1) (2,-1) and (3,2).

$(1-h)^2+(1-k)^2=r^2$                .... (1)

$(2-h)^2+(-1-k)^2=r^2$              .... (2)

$(3-h)^2+(2-k)^2=r^2$              .... (3)

Subtract equation (1) from equation (2).

$2h-4k=3$                        ..... (4)

Subtract equation (1) from equation (3).

$4h+2k=11$                      .... (5)

On solving (4) and (5) we get

$h=\frac{5}{2},k=\frac{1}{2}$

Put these values in equation (1).

$(1-\frac{5}{2})^2+(1-\frac{1}{2})^2=r^2$

$\frac{5}{2}=r^2$

The radius of the circle is

$\sqrt{\frac{5}{2}}=r$

Put the value of h,k and r in general form of the circle.

$(x-\frac{5}{2})^2+(y-\frac{1}{2})^2=\frac{5}{2}$

Therefore the equation of the circle is $(x-\frac{5}{2})^2+(y-\frac{1}{2})^2=\frac{5}{2}$.