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

None of these



easy
Solution
The point of intersection of the lines 2x – 3y + 4 = 0 and
3x + 4y – 5 = 0 is
Therefore, the centre of the circle is at
Since the origin lies on the circle, its distance from the centre of the circle is radius of the circle, therefore,
SIMILAR QUESTIONS
Find the equation of the normal to the circle x^{2} + y^{2} = 2x, which is parallel to the line x + 2y = 3.
Find the equation of the circle which cuts orthogonally each of the three circles given below:
x^{2} + y^{2} – 2x + 3y – 7 = 0, x^{2} + y^{2} + 5x – 5y + 9 = 0 and x^{2} + y^{2} + 7x – 9x + 29 = 0.
Circum centre of the triangle PT_{1}T_{2} is at
If P is taken to be at (h, 0) such that P’ lies on the circle, the area of the rhombus is
Locus of midpoint of the chords of contact of x^{2} + y^{2} = 2 from the points on the line 3x + 4y = 10 is a circle with centre P. If O be the origin then OP is equal to
Suppose ax + bx + c = 0, where a, b, c are in A.P. be normal to a family or circles. The equation of the circle of the family which intersects the circle x^{2} + y^{2} – 4x – 4y – 1 = 0 orthogonally is
Find the equation of chord of x^{2} + y^{2} – 6x + 10y – 9 = 0 which is bisected at (–2, 4).
Find the equation of that chord of the x^{2} + y^{2} = 15 which is bisected at (3, 2).
Find the centre and radius of the circle
2x^{2} + 2y^{2} = 3x – 5y + 7
Find the equation of the circle concentric with the circle x^{2} + y^{2} – 8x + 6y– 5 = 0 and passing through the point (–2, –7).