Equation of Circle with (h,k) as Center
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How do you find the centre of a circle given two points
- x2+y2+2x−2y=62
- x2+y2−2x+2y=47
- x2+y2+2x−2y=47
- x2+y2−2x+2y=62
P(x1, y1), Q(x2, y2), R(x3, y3), and S(x4, y4), then
- x1+x2+x3+x4=0
- y1+y2+y3+y4=0
- x1x2x3x4=c4
- y1y2y3y4=c4
- x2+y2=1
- x2+y2=√2
- x2+y2=4
- x2+y2=−4
- x2+y2−18x−16y−120=0
- x2+y2−18x−16y+120=0
- x2+y2+18x+16y−120=0
- x2+y2+18x−16y+120=0
- (−5, 2)
- (2, −5)
- (5, −2)
- (−2, 5)
- 4x+5y−6=0
- 2x−3y+10=0
- 3x+4y−3=0
- 5x+2y+4=0
- x2+y2−10x−6y−66=0
- x2+y2−10x−6y+100=0
- x2+y2−10x−6y+66=0
- x2+y2−10x−6y−100=0
- Hyperbola with eccentricity √2
- Hyperbola with eccentricity √3
- Ellipse with eccentricity 1√2
- Parabola with focus (0, 2a)
- x2+y2−8x+10y=0
- x2+y2+8x−10y+12=0
- x2+y2+10x−8y=0
- x2+y2+10x−8y+25=0
- a+b=18
- a+b=√2
- a−b=4√2
- a.b=73
- (x−1)2+(y−6)2=2
- (x−1)2+(y+6)2=2
- (x+1)2+(y−6)2=2
- (x+1)2+(y+6)2=2
- y=√1+2x, x≥0
- y=√1+4x, x≥0
- x=√1+4y, y≥0
- x=√1+2y, y≥0
- x2+y2+4x−10y+25=0
- x2+y2−4x−10y+25=0
- x2+y2−4x−10y+16=0
- x2+y2−14y+8=0
The circle passing through (1, -2) and touching the axis of x at (3, 0) also passes through the point
(-5, 2)
(2, -5)
(5, -2)
(-2, 5)
Let c1:x2+y2=1;C2:(x−10)2+y2=1 and C3;x2+y2−10x−42y+457=0 be three circle.A circle C has been drawn to touch circles C1 and C2 externally and C3 internally. Now circles C1, C2 and C3 start rolling on the circumference of circle C in anticlockwise direction with constant speed. The centroid of the triangle formed by joining the centres of rolling circles C1, C2 and C3 lies on
x2+y2−12x−22y+144=0
x2+y2−10x−24y+144=0
x2+y2−8x−20y+64=0
x2+y2−4x−2y−4=0
Let a, b, c, d be real numbers such that {a2+b2+2a−4b+4=0c2+d2−4c+4d+4=0. Let m and M be the minimum and the maximum values of (a−c)2+(b−d)2, respectively. The value of m×M is
(correct answer + 3, wrong answer 0)