Eq ofcircle class 11

Question

Find eq of circle passing through point (2,4) having centre at intersection of lines x -2y = 5 and 3x - y = 5

Answer 1

The standard form of an equation of circle is:  (x - h)² + (y - k)² = r²

Where (h,k) is the centre of the circle and r is the radius of the circle.

We are given that the centre of the circle is at the intersection of lines x-2y=5 and 3x-y=5

Solving for the intersection of the two lines we get: (1,-2)

Substituting those values for h and k in the standard form we get: (x - 1)² + (y - (-2))² = r²

Or: (x - 1)² + (y + 2)² = r²

To find the value of r², we use the fact that the circle passes through the point (2,4).

Substituting those values for x and y in our equation we get: (2 - 1)² + (4 + 2)² = r²

Solving for r² we get 37.

Finally we have the equation of the circle we were looking for:  (x - 1)² + (y + 2)² = 37