Equation of the Circle Using End Points of Diameter
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The equation of the circumcircle of an equilateral triangle is and one vertex of the triangle is . The equation of incircle of the triangle is
None of these
- (x−4)2+(y−3)2=7
- x2+y2=25
- x2+y2=49
- (x−3)2+(y−4)2=25
Find the coordinates of the centre of the circle passing through the points (0, 0), (–2, 1) and (–3, 2). Also, find its radius. [4 MARKS]
The radius of the circle x2 + y2 − 2x + 4y − 11 = 0
is
- -1 units
- 2 units
- 11 units
- 4 units
Consider the points A(3, 2), B(9, 4), C(7, 10). Then the circle with AC as diameter doesn't pass through B.
False
True
- 4
The equation of the circle having (−2, 1) and (4, −1) as the endpoints of its diameter is
x2+y2−2x−9=0
x2+y2−2x+9=0
x2+y2+2x+9=0
x2+y2+2x−9=0
Consider the points A(4, 3) and B(0, 1). If a circle is drawn with AB as diameter, then the coordinates of the points where this circle intersects the x-axis are
(−3, 0) and (2, 0).
(3, 0) and (−1, 0).
(3, 0) and (1, 0)
(2, 0) and (−1, 0).
A circle is drawn with AB as the diameter whose endpoints are A(2, 4) and B(4, 12). If another circle with diameter one-third of the above circle is drawn with the same centre, what are the points that the circle cuts AB?
(6, 283) and (6, 203)
(10, 283) and (6, 203)
(10, 283) and (10, 203)
(10, 203) and (6, 203)
The equation of the circle having (−2, −1) and (2, 1) as the endpoints of its diameter is
x2+y2=5
x2−y2=5
x2+y2=−5
x2−y2=−5
If the centre of a circle is (2a, a-7), then find the values of a , if the circle passes through the point (11, -9) and has diameter 10√2 units.
The equation of circle with centre (1, 1) and radius 4 is given by (x−1)2 + (y−1)2 = 4
True
False
A circle is drawn with AB as the diameter whose endpoints are A(2, 4) and B(4, 12). If another circle with diameter one-third of the above circle is drawn with the same centre, what are the points that the circle cuts AB?
(6, 283) and (6, 203)
(10, 283) and (6, 203)
(10, 283) and (10, 203)
(10, 203) and (6, 203)
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ΔPQR if C is the origin?
Also calculate the areas of the triangles in these cases. What do you observe?
The equation of the circle having (3, 3) and origin as the endpoints of its diameter is
x2+y2−3(x+y)=0
x2+y2−3x+y=0
(x+y)2−2xy−3(x+y)=0
(x+y)2+2xy−3x+y=0
- (3, -2)
- (-3, 2)
- (2, -3)
- (-2, 3)
- (x−3)2+(y−4)2=61
- (x−7)2+(y+4)2=269
- (x+7)2+(y−4)2=61
- (x+3)2+(y+4)2=169
- (x+3)2+(y−4)2=25
i) xy−x−y+1=0
ii) x2−y2−2x+2y=0
iii) x2+y2−4x+6y+3=0
The equation of the circle having (6, 7) and (4, 3) as the endpoints of its diameter is
(x+y)2−2xy+45=0
(x2−y2−2xy+45=0
10(x+y)−2xy+45=0
(x+y)[(x+y)−10]−2xy+45=0
Equation of the circle passing through (2, 3) and centre at the origin is _______ .
(x−2)2+(y−3)2=13
x2+y2=13
(x−3)2+(y−2)2=13
x2−y2=13
The equation of the circle having (4, −1) and (2, −1) as the endpoints of its diameter is
x2+y2−x+2y+9=0
x2+y2−6x+2y−9=0
x2+y2+x+y−9=0
x2+y2−6x+2y+9=0
Find the radius of the circle x2 + y2 − 2x + 4y − 11 = 0
- x2+y2=13
- x2+y2=12
- x2+y2=11
- x2+y2=1
given equation : ax2+by2+cx+dy+e=0
- (3, −5)
- (32, −4)
- (−3, −5)
- (3, 5)
The equation of the circle having (4, 2) and (2, −12) as the end points of its diameter is
x2+y2+6x+10y+16=0
x2+y2−6x+10y−16=0
x2+y2+10x−6y+16=0
x2+y2−10x+6y−16=0