Equation of Circle with (h,k) as Center
Trending Questions
Q.
Point M moved along the circle (x−4)2+(y−8)2=20. Then it broke away from it and moving along a tangent to the circle, cuts the x-axis at the point (–2, 0). The coordinates of the point on the circle at which the moving point broke away can be
(3, 5)
(6, 4)
none of these
Q. The area of the triangle whose sides are represented by the graphs of the equations y =x, x = 0 and x + y = 5, is (1) 6 sq. units(2) 6.25 sq. units, (3) 12.5 sq. units(4) 20 sq. units
Q. A square inscribed in the circle x2+y2−2x+4y+3=0. Its side are parallel to the coordinate axes. Then, one vertex of the square is
- (1+√2, 2)
- (1−√2, −2)
- (1, −2+√2)
- None of these
Q. The equation of circle of radius 5 and touching the coordinates axes in third quadrant, is
- (x+5)2+(y+5)2=25
- (x−5)2+(y−5)2=25
- (x−5)2+(y+5)2=25
- (x+5)2+(y−5)2=25
Q.
For hyperbola x225−y216=1, and circle x2+y2=100, the ray from one of concyclic points passes through a focus.
- 20√3x+(20√29+41)y+20√123=0
- 20√3x+(10√29+41)y+20√123=0
- 10√3x+(10√29+41)y+20√123=0
- 20√3x+(20√29+41)y+20√41=0
Q. Find the equation of the circle which passes through (0, 1), (1, 0) and (2, 1).
Q. The centres of a set of circle, each of radius 3, lie on the circle x2+y2=25. The locus of any point inside the set of circle is
- 4≤x2+y2≤64
- x2+y2≤25
- x2+y2≥25
- 3≤x2+y2≤9
Q. The equation of the circle whose radius is 5 and which touches the circle x2+y2−2x−4y−20=0 externally at the point (5, 5), is
- x2+y2−18x−16y−120=0
- x2+y2+18x−16y+120=0
- x2+y2+18x+16y−120=0
- x2+y2−18x−16y+120=0
Q. The area of a triangle is 5 and its two vertices are A(2, 1) and B(3, −2). The third vertex lies on y=x+3. Then third vertex is
- (0, 0)
- (52, 52)
- (−32, 32)
- (72, 132)
Q. (−6, 0), (0, 6) and (−7, 7) are the vertices of a ΔABC. The incircle of the triangle has the equation.
- x2+y2+9x−9y+36=0
- x2+y2+9x+9y−36=0
- x2+y2−9x−9y+36=0
- x2+y2+18x−18y+36=0
Q. The equation of the circle whose diameter lies on 2x + 3y = 3 and 16x - y = 4 which passes through (4, 6) is
- 5(x2+y2)−3x−8y=200
- x2+y2−4x−8y=200
- 5(x2+y2)−4x=200
- x2+y2=40
Q. The area bounded by the closed curve whose equation is x2−6x+y2+8y=0 is
- 12π
- 25π
- 36π
- cannot be determined
- 48π
Q. If the points (2 , 0) , (0 , 1) , (4 , 5) and (0 , C) are concyclic , then the value of C is
- 1
- 143
- None of these
- 5