Question

Centres of the three circles
$x^2+y^2-4x-6y-14=0$
$x^2+y^2+2x+4y-5=0$ and
$x^2+y^2-10x-16y+7=0$. The centres of the circles are:

• A. are the vertices of a right angle
• B. the vertices of an isosceles triangle which is not regular
• C. vertices of a regular triangle
• D. are collinear

Answer 1

Answer: (D)

are collinear

$x^2+y^2-4x-6y-14=0$ ---(i)

$x^2+y^2+2x+4y-5=0$ ---(ii)

$x^2+y^2-10x-16y+7=0$ ---(iii)

centre of the circle (i) $c_1=\left(-g,-f\right)=\left(2,3\right)$

centre of the circle (ii) $c_2=\left(-g,-f\right)=\left(-1,-2\right)$

centre of the circle (iii) $c_3=\left(-g,-f\right)=\left(5,8\right)$

Now, $c_1{c}_2=\sqrt{{\left(2+1\right)}^2+{\left(3+2\right)}^2}=\sqrt{34}$

$c_2{c}_3=\sqrt{{\left(5+1\right)}^2+{\left(8+2\right)}^2}=2\sqrt{34}$

$c_1{c}_3=\sqrt{{\left(5-2\right)}^2+{\left(8-3\right)}^2}=\sqrt{34}$

Since $c_1{c}_2+c_1{c}_3=c_2{c}_3$

Therefore $c_1$,$c_2$ and $c_3$ collinear.