Question

Find the equation of the circle which passes through the points $A(1,1)$ and $B(2,2)$ and whose radius is $1$. Show that there are two such circles.

Answer 1

Let C is the center of circle with coordinates $C(h,k)$

Coordinates of A are $A(1,2)$

Coordinates of point B are $B(2,2)$

As points A and B lie on circle, AC and BC are radii of circle.

By distance formula,

$AC=r=\sqrt{{(h-1)}^2+{(k-2)}^2}$

$\therefore r^2={(h-1)}^2+{(k-2)}^2$

$\therefore {(h-1)}^2+{(k-2)}^2=1$ Equation (1)

Similarly, $BC=r=\sqrt{{(h-2)}^2+{(k-2)}^2}$

$\therefore r^2={(h-2)}^2+{(k-2)}^2$

$\therefore {(h-2)}^2+{(k-2)}^2=1$ Equation (2)

Perform Equation (1) - Equation (2),

${(h-1)}^2-{(h-2)}^2=0$

$\therefore (h^2-2h+1)-(h^2-4h+4)=0$

$\therefore h^2-2h+1-h^2+4h-4=0$

$\therefore 2h-3=0$

$\therefore h=\frac{3}{2}$

Put this value in equation (2),

${(\frac{3}{2}-2)}^2+{(k-2)}^2=1$

${\left(\frac{-1}{2}\right)}^2+k^2-4k+4=1$

$\frac{1}{4}+k^2-4k+4=1$

$k^2-4k+\frac{13}{4}=0$

Multiply both sides by 4, we get,

$4k^2-16k+13=0$

$\therefore k=\frac{-(-16)\pm \sqrt{{(-16)}^2-4\times 4\times 13}}{2\times 4}$

$\therefore k=\frac{16\pm \sqrt{256-208}}{8}$

$\therefore k=\frac{16\pm \sqrt{48}}{8}$

$\therefore k=\frac{16\pm 4\sqrt{3}}{8}$

$\therefore k=\frac{4(4\pm \sqrt{3})}{8}$

$\therefore k=\frac{(4\pm \sqrt{3})}{2}$

$\therefore k=\frac{(4+\sqrt{3})}{2}$ and $\therefore k=\frac{(4-\sqrt{3})}{2}$

Thus, there are two centers of circle i.e. $C_1(\frac{3}{2},\frac{(4+\sqrt{3})}{2})$ and $C_1(\frac{3}{2},\frac{(4-\sqrt{3})}{2})$

Thus, there are two such circles.