Obtaining Centre and Radius of a Circle from General Equation of a Circle
Trending Questions
Q. A square is inscribed in the circle x2+y2−2x+4y+3=0, whose sides are parallel to the coordinate axes. One vertex of the square is
- (1+√2, −2)
- (1−√2, −2)
- (1−2+√2)
- None of these
Q. Centre of the circle (x−3)2+(y−4)2 = 5 is
- (3, 4)
- (-3, -4)
- (4, 3)
- (-4, -3)
Q.
An infinite number of tangents can be drawn from (1, 2) to the circle x2+y2−2x−4y+λ=0, then λ=
-20
0
5
Cannot be determined
Q. If g2+f2=c, then the equation x2+y2+2gx+2fy+c=0 will represent
A circle of radius g
A circle of radius f
A circle of diameter √c
- A circle of radius 0
Q.
The number of integral values of λ for which x2+y2+λx+(1−λ)y+5=0 is the equation of a circle whose radius cannot exceed 5, is
- 14
- 18
- 16
- none of these
Q. The radius of circle x2+y2−12x+8y+32=0, is
- 2√5
- √10
- 4√11
- √5
Q. The circle passing through the intersection of the circles, x2+y2−6x=0 and x2+y2−4y=0, having its centre on the line, 2x−3y+12=0, also passes through the point
Q. If the circle x2+y2−6x−10y+c=0 does not touch or intersect the coordinate axes and (1, 4) lies inside the circle, then the number of integral values of c is
Q.
The coordinates of the centre of a circle, whose radius is 2 units and which touches the line pair x2−y2−2x+1=0, are
- (4, 0)
- (1+2√2, 0)
- (4, 1)
- (1, 2√2)
Q. If kx2+ky2−4x−6y−2k=0 represents a real circle, then the set of values of k is
- (0, ∞)
- (−∞, 0)
- R
- R−{0}
Q. The equation of the circle passsing through (1, 0) and (0, 1) and having the smallest possible radius is
- x2+y2+2x+2y=0
- x2+y2−x−y=0
- x2+y2+x+y=0
- x2+y2−2x−2y=0
Q.
___
Find the radius of the circle x2 + y2 − 2x + 4y − 11 = 0
Q. If the two circles (x−1)2+(y−3)2=r2 and x2+y2−8x+2y+8=0 intersect at two distinct points, then
- 2 < r < 8
- r < 2
- r = 2
- r > 2
Q. Let C be a circle passing through the origin and making an intercept of √10 on the line y=2x+5√2. If the line subtends an angle of 45∘ at the origin, then the equation of circle C is/are
- x2+y2−4x−2y=0
- x2+y2−2x−4y=0
- x2+y2+4x+2y=0
- x2+y2+2x+4y=0
Q. If the line x + 2by + 7 = 0 is diameter of the circle x2+y2−6x+2y=0, then b =
- 3
- -5
- -1
- 5
Q. If 2x2+2y2−12x+8y+k=0 is a point circle, then the value of k is
- 18
- 12
- 13
- 26
Q. The equation of the image of circle x2+y2−4x+6y+10=0 in the line 4x−3y+8=0 is
- (x−6)2+(y−3)2=3
- (x+6)2+(y−3)2=3
- (x+6)2+(y+3)2=3
- (x−6)2+(y+3)2=3