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
A.
(x-p)2 = 4qy
In a circle, AB is a diameter where the co-ordinate of A is (p, q) and let the co-ordinate of B is (x1 , y1 ).
Equation of circle in diameter form is (x - p)(x - x1 ) + (y - q)(y - y1 ) = 0
x2 - (p + x1 )x + px1 + y2 - (y1 + q)y + qy1 = 0
x2 - (p + x1 )x + y2 - (y1 + q)y + px1 + qy1 = 0
Since this circle touches X-axis
∴ y = 0
⇒ x2 - (p + x1 )x + px1 + qy1 = 0 Also the discriminant of above equation will be equal to zero because circle touches X-axis.
∴ (p + x1 )2 = 4(px1 + qy1) p2 + x21 + 2px1
= 4px1 + 4qy1 x21 - 2px1 + p2 = 4qy1
Therefore the locus of point B is, (x - p)2 = 4qy
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