Find the power of the point (2, 4) with respect to the circle. x2 + y2 – 6x+ 4y – 8 = 0
The power of point (2, 4) w.r.to the circle x2 + y2 – 6x – 4y – 8 = 0 is
Let S ≡ x2 + y2 + 2gx + 2fy + c = 0 be a given circle. find the locus of the foot of the perpendicular drawn from the origin upon any chord of S which subtends a right angle at the origin.
The normal 3x – 4y = 4 and 6x – 8y – 7 = 0 are tangents to the circle. Then its radius is:
The circle x2 + y2 + x + y = 0 and x2 + y2 + x – y = 0 intersect at the angle of:
Find the radical centre of the circles, x2 + y2 + 3x + 2y + 1 = 0, x2 + y2 – x + 6y + 5 = 0, x2 + y2 + 5x – 8y + 15 = 0
The tangents drawn from the origin to the circle x2 + y2 – 2kx – 2ry + r2 = 0 are perpendicular, if:
The locus of the mid points of a chord of the circles x2 + y2 = 4, which subtends a right angle at the origin is:
The chord of contact of tangents from a point P to a circle passes through Q, if l1 and l2 are the lengths of tangents from P and Q to the circle, then PQ is equal to:
Find the equation of the chord of x2 + y2 – 6x + 10y – 9 = 0 which is bisected at (–2, 4).
Find the middle point of the chord intercepted on line lx + my + n = 0 by the circle x2 + y2 = a2.
Tangent which is parallel to the line x – 3y – 2 = 0 of the circle x2 + y2 – 4x + 2y – 5 = 0, has point/points of contact.