The equation of sphere concentric with the sphere x2 + y2 + z2 - 4x - 6y - 8z - 5 = 0 and which passes through the origin, is
x2 + y2 + z2 - 4x - 6y - 8z = 0
x2 + y2 + z2 - 6y - 8z = 0
x2 + y2 + z2 = 0
x2 + y2 + z2 - 4x - 6y - 8z - 6 = 0
If two circles 2x2 + 2y2 - 3x + 6y + k = 0 and x2 + y2 - 4x + 10y + 16 = 0 cut orthagobally then, value of k is
41
14
4
1
C.
4
We have given two circles are
2x2 + 2y2 - 3x + 6y + k = 0
Since, general equation of circle is
x2 + y2 + 2gx + 2fy + c = 0 ...(iii)
Therefore, comparing Eqs. (i) and (ii) with Eq. (iii), we get
The locus of z satisfying the inequality , where z = x + iy, is
x2 + y2 < 1
x2 - y2 < 1
x2 + y2 > 1
2x2 + 3y2 < 1
The area (in square unit) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is
4
The equations of the circle which pass through the origin and makes intercepts of lengths 4 and 8 on the x and y -axes respectively are
x2 + y2 ± 4x ± 8y = 0
x2 + y2 ± 2x ± 4y = 0
x2 + y2 ± 8x ± 16y = 0
x2 + y2 ± x ± y = 0
The point (3 -4) lies on both the circles x2 + y2 - 2x + 8y + 13 = 0 and x2 + y2 - 4x + 6y + 11 = 0. Then, the angle between the circles is
135
The equation of the circle which passes through the origin and cuts orthogonally each of the circles x2 + y2 - 6x + 8 = 0 and x2 + y2 - 2x - 2y = 7 is
3x2 + 3y2 - 8x - 13y = 0
3x2 + 3y2 - 8x + 29y = 0
3x2 + 3y2 + 8x + 29y = 0
3x2 + 3y2 - 8x - 29y = 0
If the circle x2 + y2 = a intersects the hyperbola xy = c2 in four points (xi, yi), for i = 1, 2, 3 and 4, then y1 + y2 + y3 + y4 equals
0
c
a
c4