Advertisements
Advertisements
प्रश्न
The equation of the circle with centre on the x axis and passing through the origin is:
पर्याय
x2 – 2ax + y2 = 0
y2 – 2ay + x2 = 0
x2 + y2 = a2
x2 – 2ay + y2 = 0
उत्तर
x2 – 2ax + y2 = 0
APPEARS IN
संबंधित प्रश्न
Find the centre and radius of the circle
x2 + y2 – 22x – 4y + 25 = 0
Find the equation of the circle on the line joining the points (1, 0), (0, 1), and having its centre on the line x + y = 1.
If the lines x + y = 6 and x + 2y = 4 are diameters of the circle, and the circle passes through the point (2, 6) then find its equation.
Determine whether the points P(1, 0), Q(2, 1) and R(2, 3) lie outside the circle, on the circle or inside the circle x2 + y2 – 4x – 6y + 9 = 0.
If the centre of the circle is (-a, -b) and radius is `sqrt("a"^2 - "b"^2)` then the equation of circle is:
Determine whether the points (– 2, 1), (0, 0) and (– 4, – 3) lie outside, on or inside the circle x2 + y2 – 5x + 2y – 5 = 0
Find centre and radius of the following circles
x2 + (y + 2)2 = 0
Find centre and radius of the following circles
x2 + y2 + 6x – 4y + 4 = 0
Choose the correct alternative:
The length of the diameter of the circle which touches the x -axis at the point (1, 0) and passes through the point (2, 3)
Choose the correct alternative:
The radius of the circle passing through the points (6, 2) two of whose diameter are x + y = 6 and x + 2y = 4 is
