The equation of the circle with center (2, 1) and touching the line 3x + 4y = 5 is 1) x2 + y2 - 4x - 2y + 5 = 0 2) x2 + y2 - 4x - 2y - 5 = 0 3) x2 + y2 - 4x - 2y + 4 = 0 4) x2 + y2 - 4x - 2y - 4 = 0 Solution: (3) x2... View Article
One of the diameter of the circle x2 + y2 – 12x + 4y + 6 = 0 is given by 1) x + y = 0 2) x + 3y = 0 3) x = y 4) 3x + 2y = 0 Solution: (2) x + 3y = 0 Given the equation of circle is x2 + y2 - 12x + 4y + 6... View Article
The other end of the diameter through the point (-1, 1) on the circle x2 + y2 – 6x + 4y – 12 = 0 is 1) (-7, 5) 2) (-7, - 5) 3) (7, - 5) 4) (7, 5) Solution: (3) (7, - 5)... View Article
The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines x = 0 and y = 0, is 1) x2 + y2 - 6x + 6y + 9 = 0 2) x2 + y2 - 6x - 6y + 9 = 0 3) x2 + y2 + 6x - 6y + 9 = 0 4) x2 + y2 + 6x + 6y + 9 = 0 Solution: (1) x2... View Article
The equation of two circles which touch the Y – axis at (0, 3) and make an intercept of 8 units on X – axis , are 1) x2 + y2 ± 10x - 6y + 9 = 0 2) x2 + y2 ± 6x - 10y + 9 = 0 3) x2 + y2 - 8x ± 10y + 9 = 0 4) x2 + y2 - 10x ± 6y + 9 = 0 Solution: (1)... View Article
A variable circle passes through the fixed point A (p, q) and touches X – axis. The locus of the other end of the diameter through A is 1) (x - p )2 = 4qy 2) (x - q )2 = 4py 3) (y - p )2 = 4qx 4) (y - q )2 = 4px Solution: (1) (x - p )2 = 4qy Since, the coordinates of one end... View Article
The area of the circle centred at (1, 2) and passing through (4, 6) is 1) 5 π sq units 2) 10 π sq units 3) 25 π sq units 4) None of these Solution: (3) 25 π sq units The centre of the circle is C (1,... View Article
The center of the circle whose normals are x2 – 2xy – 3x + 6y = 0, is 1) (3, 3 / 2) 2) (3, - 3 / 2) 3) (3 / 2, 3) 4) None of these Solution: (1) (3, 3 / 2) x2 - 2xy - 3x + 6y = 0 (x - 3) (x - 2y) = 0... View Article
If the line y = 7x – 25 meets the circle x2 + y2 = 25 in points A, B, then the distance between A and B is 1) √10 2) 10 3) 5√2 4) 5 Solution: (3) 5√2 y = 7x – 25 …(i) x2 + y2 = 25 …(ii) Put (i) in (ii) => x2 + (7x – 25)2 = 25... View Article
The locus of the centre of a circle, which touches externally the circle x2 + y2 – 6x – 6y + 14 = 0 and also touches the y-axis, is given by the equation: 1) x2 – 6x – 10y + 14 = 0 2) x2 – 10x – 6y + 14 = 0 3) y2 – 6x – 10y + 14 = 0 4) y2 – 10x – 6y + 14 = 0 Solution: (4) y2 – 10x – 6y +... View Article
The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle x2 + y2 = 9 is 1) (3 / 2, 1 / 2) 2) (1 / 2, 3 / 2) 3) (1 / 2, 1 / 2) 4) (1 / 2, - 21/2) Solution: (4) (1 / 2, - 21/2) x2 + y2 + 2gx + 2fy + c = 0... View Article
The equation of the smallest circle passing through the points (2, 2) and (3, 3) is 1) x2 + y2 + 5x + 5y + 12 = 0 2) x2 + y2 - 5x - 5y + 12 = 0 3) x2 + y2 + 5x - 5y + 12 = 0 4) x2 + y2 - 5x + 5y - 12 = 0 Solution: (2) x2 + y2... View Article
If one end of the diameter is (1, 1) and the other end lies on the line x + y = 3, then locus of centre of circle is 1) x+ y = 1 2) 2(x - y ) = 5 3) 2x + 2y = 5 4) None of these Solution: (3) 2x + 2y = 5... View Article
If the lines 2x – 3y = 5 and 3x – 4y = 7 are two diameters of a circle of radius 7, then the equation of the circle is 1) x2 + y2 + 2x - 4y - 47 = 0 2) x2 + y2 = 49 3) x2 + y2 - 2x + 2y - 47 = 0 4) x2 + y2 = 17 Solution: (3) x2 + y2 - 2x + 2y - 47 = 0... View Article
The radius of the circle with the polar equation r2 – 8r(√3 cos θ + sin θ) + 15 = 0 is 1) 8 2) 7 3) 6 4) 5 Solution: (2) 7... View Article
The point diametrically opposite to the point P(1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is 1) (3, 4) 2) (3, - 4) 3) (-3, 4) 4) (-3, - 4) Solution: (4) (-3, - 4) The centre of the given circle is (-1, -2). Let another end... View Article
If the tangent at point P on the circle x2 + y2 + 6x + 6y – 2 = 0 meets the straight line 5x – 2y + 6 = 0 at a point Q on y – axis, the length of PQ is 1) 4 2) 2√5 3) 5 4) 3√5 Solution: (3) 5 The line 5x - 2y + 6 = 0 meets the y-axis at the point (0, 3) and therefore the tangent has to pass... View Article
If the equation λx2 + (2λ – 3)y2 – 4x – 1 = 0 represents a circle, then its radius is 1) √11/3 2) √13/3 3) √7/3 4) 1/3 Solution: (3) √7/3... View Article
The equation of circle which touches X and Y – axes at the points (1, 0) and (0, 1) respectively is 1) x2 + y2 - 4y + 3 = 0 2) x2 + y2 - 2y - 2 = 0 3) x2 + y2 - 2x - 2y + 2 = 0 4) x2 + y2 - 2x - 2y + 1 = 0 Solution: (4) x2 + y2 - 2x... View Article
The diameters of a circle are along 2x + y – 7 = 0 and x + 3y – 11 = 0. Then, the equation of this circle, which also passes through (5, 7), is 1) x2 + y2 - 4x - 6y - 16 = 0 2) x2 + y2 - 4x - 6y - 20 = 0 3) x2 + y2 - 4x - 6y - 12 = 0 4) x2 + y2 + 4x + 6y - 12 = 0 Solution: (3) x2 + y2... View Article
