The graph of the function y = f(x) is symmetrical about the line x = 2, then
f(x + 2)= f(x – 2)
f(2 + x) = f(2 – x)
f(x) = f(-x)
f(x) = f(-x)
If 2a + 3b + 6c =0, then at least one root of the equation ax2 + bx+ c = 0 lies in the interval
(0,1)
(1,2)
(2,3)
(2,3)
If the sum of the slopes of the lines given by x2 -2cxy -7y2 =0 is four times their product, then c has the value
-1
2
-2
-2
The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is –1 is
If a circle passes through the point (a, b) and cuts the circle x2 +y2= 4 orthogonally, then the locus of its centre is
2ax +2by + (a2 +b2+4)=0
2ax +2by - (a2 +b2+4)=0
2ax -2by - (a2 +b2+4)=0
2ax -2by - (a2 +b2+4)=0
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
(x-p)2 = 4qy
(x-q)2 = 4py
(y-p)2 = 4qx
(y-p)2 = 4qx
If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10π, then the equation of the circle is
x2 + y2- 2x +2y -23 = 0
x2 - y2- 2x -2y -23 = 0
x2 - y2- 2x -2y +23 = 0
x2 - y2- 2x -2y +23 = 0
A.
x2 + y2- 2x +2y -23 = 0
The lines 2x + 3y + 1 = 0 and 3x - y - 4 = 0 are diameters of circle.
On solving these equations we get x = 1, y = -1
Therefore the centre of circle = (1, -1) and circumference = 10 π
2πr = 10π
⇒ r = 5
∴ Equation of circle (x - x1 )2 + (y - y1 )2 = r2
(x - 1)2 + (y + 1)2 = 52
x2 + 1 - 2x + y2 + 2y + 1 = 25
x2 + y2 - 2x + 2y - 23 = 0