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  • Find the equation of the circle passing through the points <img alt="\left ( 3,4 \right )\: and\; \left ( 5,6 \right )" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cleft%20%28%203%2C4%20

Find the equation of the circle passing through the points \left ( 3,4 \right )\: and\; \left ( 5,6 \right ) and whose centre is on the line x+2y= 7

Option: 1

\left ( x-3 \right )^{2}+\left ( y-5 \right )^{2}= 8


Option: 2

\left ( x-4 \right )^{2}+\left ( y-5 \right )^{2}= 9


Option: 3

\left ( x-2 \right )^{2}+\left ( y-3 \right )^{2}= 10


Option: 4

\left ( x-5 \right )^{2}+\left ( y-7 \right )^{2}= 6


Answers (1)

best_answer

Let the equation of the circle be \left ( x-a \right )^{2}+\left ( y-b \right )^{2}= r^{2}

Since the circle passes through \left ( 3,4 \right )\: and\: \left ( 5,6 \right ), we have the following two equations:

\left ( 3-a \right )^{2}+\left ( 4-b \right )^{2}= r^{2}\cdots \left ( 1 \right )
\left ( 5-a \right )^{2}+\left ( 6-b \right )^{2}= r^{2}\cdots \left ( 2 \right )

Solving equations (1) and (2), we get a= 4\: and\: b= 5
Now, the centre of the circle lies on the line x+2y= 7.

Substituting a= 4\: and\: b= 5 in x+2y= 7, we get 4+2\left ( 5 \right )= 14.
Thus, the equation of the circle is \left ( x-4 \right )^{2}+\left ( y-5 \right )^{2}= r^{2}.
Substituting \left ( 3,4 \right ) in the equation, we get r^{2}= 5. Substituting this value of r^{2}  in the equation, we get \left ( x-4 \right )^{2}+\left ( y-5 \right )^{2}= 9 as the correct answer.

Posted by

Irshad Anwar

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