A circle whose centre is the point of intersection of the lines 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 passes through the origin. Find its equation.

Given that the circle has the centre at the intersection point of the lines 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0 and passes through the origin.



Let us find the points of intersection of the lines.


On solving the lines 2x - 3y + 4 = 0 and 3x + 4y - 5 = 0, we get the point of intersection to be


We have circle with centre and passing through the point (0,0).


We know that radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.


We know that the distance between the two points (x1,y1) and (x2,y2) is .


Let us assume r is the radius of the circle.






..... (1)


We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:


(x - p)2 + (y - q)2 = r2


Now we substitute the corresponding values in the equation:





17x2 + 17y2 + 2x - 44y = 0.


The equation of the circle is 17x2 + 17y2 + 2x - 44y = 0.


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