Question

Cantres of the three circles
$x^2+y^2-4x-6y-14=0$
$x^2+y^2+2x+4y-5=0$
and $x^2+y^2-10x-16y+7=0$

• A. Are the vertices of a right triangle
• 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

Centre of $x^2+y^2-4x-6y-14=0is(-2,-3)$

Centre of $x^2+y^2+2x+4y-5=0is(1,2)$

Centre of $x^2+y^2-10x-16y+7=0is(-5,-8)$

Let these centres be $A(-2,-3);B(1,2);C=(-5,-8)$

Now we find the lengths of $AB,BC,CA$

$\Rightarrow AB=\sqrt{(-2-1{)}^2+(-3-2{)}^2}=\sqrt{34}$

$\Rightarrow BC=\sqrt{(-5-1{)}^2+(-8-2{)}^2}=\sqrt{136}$

$\Rightarrow CA=\sqrt{(-2+5{)}^2+(-3+8{)}^2}=\sqrt{34}$

Here we can see that $AB+CA=BC$

Therefore $ABC$ is a straight line.