Question

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

• 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

The correct option is D Are collinear

From the given equation, we get the centers $A(2,3),B(-1,-2),C(5,8)$, respectively.

Using the distance formula as written below-

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

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

$AB=\sqrt{9+25}$

$AB=\sqrt{34}unit$

Similarly,

$AC=\sqrt{{(2-5)}^2+{(3-8)}^2}$

$AC=\sqrt{9+25}$

$AC=\sqrt{34}unit$

Similarly,

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

$BC=\sqrt{36+100}$

$BC=\sqrt{136}=2\sqrt{34}unit$

Since, $AB+AC=BC$

Hence, the points are collinear.