Obtaining Centre and Radius of a Circle from General Equation of a Circle
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Find the equation of chord of contact of the point (3, 5) on the circle x2+y2+12x+8y+7=0.
9x - y - 5 = 0
9x + y + 5 = 0
13x - 2y +1 =0
13 x + 2y - 1 = 0
- p2=q2
- p2=8q2
- p2<8q2
- p2>8q2
- 2
- 4
Find the centre of a circle passing through points and
through the origin, then the equations of OA and OB are
- 3x – y = 0, x + 3y = 0
- 3x – y = 0, x – 3y = 0
- 3x + y = 0, x – 3y = 0
- 3x + y = 0, x + 3y = 0
- (2, -3)
- (-2, 3)
- (3, -2)
- (-3, 2)
If two circles (x−1)2+(y−3)2=r2 and x2+y2−4x−4y+4=0 intersect in two distinct point then
2 < r < 8
r < 2
r = 2
r > 2
The centre and radius of the circle x2 + y2 + 2gx + 2yf + c=0 are
(g, f) and √g2 + f2 − c
(−g, −f) and √g2 + f2 − c
(g, f) and √g2 + f2 + c
(−g, −f) and √g2 + f2 + c
(2 + 2λ + μ)x2 + (1 + 2λ + 2μ)y2 + (3 + 5λ + 3μ)xy
− (16 + 14λ + 11μ)x − (10 + 13λ + 17μ) + (24 + 20λ + 30μ) = 0 is an equation of circle only when
λ=1 μ=1
λ=−2 μ=−1211
λ=1 μ=0
λ=−65 μ=1
The area of the circle whose centre is at (1, 2) and which passes through the point (4, 6) is