Question

The centre of the circle passing through the points $(8,12),(11,3)$ and $(0,14)$ is

• A. $(2,5)$
• B. $(-2,5)$
• C. $(3,5)$
• D. $(4,6)$

Answer 1

The correct option is A $(2,5)$

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ............(i)

Then its centre will be $O=(-g,-f)$.

It passes through $A=(x_1,y_1)=(8,12),B=(x_2,y_2)=(11,3)$ and $C=(x_3,y_3)=(0,14)$.

Substituting $(x,y)=(x_1,y_1)=(8,12),(x_2,y_2)=(11,3)$ and $(x_3,y_3)=(0,14)$ in equation (i) and simplifying, we get

$16g+24y+c=-208$ ...........(ii),

$22g+6f+c=-130$ .............(iii)

and $28f+c=-196$ ...............(iv)

Eliminating $c$ from (ii) and (iii),

$g=3f+13$ ..........(v)

Again eliminating $c$ from (iii) and (iv), we have

$g=f+3$ ......(vi)

$\therefore$ From (v) and (vi), we get

$3f+13=f+3$

$\Rightarrow f=-5$

Putting $f=-5$ in (vi), have

$g=-2$

$\therefore$ Centre $O=(-g,-f)=(2,5)$