Question

Solution

Correct option is

4a2 + 4b2 – c2 – 8ab + 4bc – 4ca < 0

Chords are bisected on the line y = x. Let (x1x1) be the mid point of the chord, then equation of the chord is T = S1   This chord passes through (ab + c  Which is quadratic in x1, since it is given that two chords are bisected on the line y = x, then x1 must have two real roots.

B2 – 4AC > 0   Hence the condition on abc is

4a2 + 4b2 – c2 – 8ab + 4bc – 4ca < 0

SIMILAR QUESTIONS

Q1

The circle x2 + y2 = 1 cuts the x-axis at P and Q. another circle with centre at Q and variable radius intersects the first circle at R above the x-axis and the line segment PQ at S. Find the maximum area of the triangleQSR.

Q2

Find the equation of a circle having the lines x2 + 2xy + 3x + 6y = 0 as its normals and having size just sufficient to contain the circle

x(x – 4) + y(y – 3) = 0.

Q3

Find the equation of the circle whose radius is 5 and which touches the circle

x2 + y2 – 2x – 4y – 20 = 0 at the point (5, 5).

Q4

Find the locus of the mid point of the chord of the circle x2 + y2 = a2which subtend a right angle at the point (pq).

Q5

Let a circle be given by

2x (x – a) + y(2y – b) = 0            (a ≠ 0, b ≠ 0)

Find the condition on a and b if two chords, each bisected by the x-axis, can be drawn to the circle from (ab/2).

Q6

The centre of the circle S = 0 lie on the line 2x – 2y + 9 = 0 and S = 0 cuts orthogonally the circle x2 + y2 = 4. Show that circle S = 0 passes through two fixed points and find their co-ordinates.

Q7 be a given circle. Find the locus of the foot of perpendicular drawn from origin upon any chord of Swhich subtends a right angle at the origin.

Q8 be a given circle. Find the locus of the foot of perpendicular drawn from origin upon any chord of Swhich subtends a right angle at the origin.

Q9

P is a variable on the line y = 4. Tangents are drawn to the circle x2 + y2= 4 from P to touch it at A and B. The perpendicular PAQB is completed. Find the equation of the locus of Q.

Q10

Find the limiting points of the circles

(x2 + y2 + 2gx + c) + λ(x2 + y2 + 2fy + d) = 0