Single Choice

The equation of the circle circumscribing the triangle formed by the lines $$x+y=6$$, $$2x+y=4$$ and $$x+2y=5$$ is:

A$$x^2+y^2+17x+19y-50=0$$
B$$x^2+y^2-17x-19y-50=0$$
C$$x^2+y^2+17x-19y-50=0$$
D$$x^2+y^2-17x-19y+50=0$$
Correct Answer

Solution

Solving given three line the vertices of the triangle are $$A(1,2),B(-2,8)$$ and $$C(7,-1)$$

And let $$P(a,b)$$ the center of the circle.
Therefore,

$$PA^2=PB^2=PC^2$$

$$\Rightarrow (a-1)^2+(b-2)^2=(a+2)^2+(b-8)^2=(a-7)^2+(b+1)^2$$

Solving above equations we get $$a=\dfrac{17}{2}$$ and $$b=\dfrac{19}{2}$$

Now radius of the circle is $$PA=\sqrt{(a-1)^2+(b-2)^2}=\sqrt{\dfrac{15^2}{4}+\dfrac{15^2}{4}}$$

Hence equation of required circle is,

$$\left(x-\dfrac{17}{2}\right)^2+\left(y-\dfrac{19}{2}\right)^2=PA^2=\dfrac{15^2}{4}+\dfrac{15^2}{4}$$

$$\Rightarrow x^2+y^2-17x-19y+50=0$$

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