Question

Find the coordinates of the centre of a circle which passes through the points A(1,2), B(3,-4) and C(5,-6).

• A. $(11,2)$
• B. $(11,-2)$
• C. $(-11,2)$
• D. $(-11,-2)$

Answer 1

The correct option is A

$(11,2)$

Let the coordinates of the centre of the circle be O$(a,b)$

Since the circle passes through the 3 points A(1,2). B(3,-4) and C(5,-6), their distance from the centre will be equal to the radius.

Therefore $O{A}^2=O{B}^2=O{C}^2$

Distance Formula = $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$O{A}^2=(a-1{)}^2+(b-2{)}^2$ ............(i)

$O{B}^2=(a-3{)}^2+(b+4{)}^2$...........(ii)

$O{C}^2=(a-5{)}^2+(b+6{)}^2$...............(iii)

Equating (i) and (ii) we get

$(a-1{)}^2+(b-2{)}^2$
= $(a-3{)}^2+(b+4{)}^2$

$\Rightarrow$ $a^2+1-2a+b^2+4-4b$
= $a^2+9-6a+b^2+16+8b$

$\Rightarrow$ $-2a+5-4b$ = $-6a+25+8b$

$\Rightarrow$ $4a-12b$ = $20$ ..........(iv)

Equating (ii) and (iii) we get

$(a-3{)}^2+(b+4{)}^2$
= $(a-5{)}^2+(b+6{)}^2$

$\Rightarrow$ $a^2+9-6a+b^2+16+8b$
= $a^2+25-10a+b^2+36+12b$

$\Rightarrow$ $4a-4b=36$ ........(v)

Solving (iv) and (v) we get $a=11,b=2$

Therefore, centre of the circle is $(11,2)$.