Obtaining Centre and Radius of a Circle from General Equation of a Circle
Trending Questions
Q. Let A be the centre of the circle x2+y2−2x−4y−20=0. Suppose that the tangents at the points B(1, 7) and D(4, −2) on the circle meet at point C. Then the area of the quadrilateral ABCD is
Q. The area of the circle whose centre is at (1, 2) and which passes through the point (4, 6) is
- 5π
- 10π
- 25π
- None of these
Q. The centres of the circles x2+y2=1, x2+y2+6x−2y=1 and x2+y2−12x+4y=1 are
- Non-collinear
- Same
- Collinear
- None of these
Q. If the radius of the circle x2+y2−18x+12y+k=0 be 11, then k =
- 4
- -4
- 49
- 347
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. The equation x2+y2+4x+6y+13=0 represents
Circle
Pair of concurrent straight lines
Pair of coincident straight lines
- Point
Q. The area of the curve x2+y2=2ax is
- πa2
- 2πa2
- 4πa2
- 12πa2
Q. The equation x2+y2=0 denotes
A point
- y-axis
A circle
- x-axis
Q. The radius of the circle x2+y2+4x+6y+13=0 is
- 0
- √26
- √23
- √13
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
