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  • Find the equation of the circle passing through the points <img alt="\left ( 1,3 \right ),\left ( 4,6 \right )\, and\, \left ( 7,3 \right )" src="https://entrancecorner.oncodecogs.com/gif.late

Find the equation of the circle passing through the points \left ( 1,3 \right ),\left ( 4,6 \right )\, and\, \left ( 7,3 \right ) whose centre is on line \mathrm{2x+y=6}.

Option: 1

x^2+y^2-4 x-4 y+4=0


Option: 2

x^2+y^2-4 x-4 y-4=0


Option: 3

x^2+y^2+4 x-4 y+4=0


Option: 4

x^2+y^2+4 x-4 y-4=0


Answers (1)

best_answer

We first need to find the centre and radius of the circle passing through the three given points. Let (h, k) be the centre of the circle. Then we have:

\begin{aligned} & (h-1)^2+(k-3)^2=r^2 \ldots(1)(u \operatorname{sing}(1,3)) \\ & (h-4)^2+(k-6)^2=r^2 \ldots \ldots(2)(u \operatorname{using}(4,6)) \\ & (h-7)^2+(k-3)^2=r^2 \ldots \ldots(3)(u \operatorname{using}(7,3)) \end{aligned}

Subtracting equation (1) from (2) and (3), we get:

\begin{aligned} & -6 h-6 k+42=0 .....\left ( 4 \right ) \\ & -2 h-2 k+26=0 .....\left ( 5 \right ) \end{aligned}

Solving equations (4) and (5), we get:

h=3, k=0

Substituting \mathrm{\left ( h,k \right )=\left ( 3,0 \right )} in equation (1), we get: \mathrm{r^{2}=10}

Therefore, the centre of the circle is (3,0) and the radius is \sqrt{10}.

Since the centre of the circle lies on the line \mathrm{2x+y=6}, we have:\mathrm{2\left ( 3 \right )+0=6}

Therefore, the centre of the circle lies on the given line.

Substituting (h, k, r)=(3,0, \sqrt{10}) in the equation of a circle, we get:(x-3)^2+y^2=10

Expanding this equation, we get:

x^2+y^2-6 x=-1

Therefore, the equation of the circle passing through the points \left ( 1,3 \right ),\left ( 4,6 \right )\, and\, \left ( 7,3 \right )  and whose centre is on line 2 x+y=6 \text { is } x^2+y^2-6 x=-1 \text {. }

The correct option is (b) x^2+y^2-4 x-4 y-4=0 .

 

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Gaurav

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