Advertisements
Advertisements
प्रश्न
Find the centre and radius of the circle.
(x + 2) (x – 5) + (y – 2) (y – 1) = 0
उत्तर
Equation of the circle is (x + 2) (x – 5) + (y – 2) (y – 1) = 0
x2 – 3x – 10 + y2 – 3y + 2 = 0
x2 + y2 – 3x – 3y – 8 = 0
Comparing this with x2 + y2 + 2gx + 2fy + c = 0
We get 2g = -3, 2f = -3, c = -8
g = `(-3)/2,` f = `(-3)/2`, c = - 8
Centre (-g, -f) = `(3/2, 3/2)`
Radius = `sqrt("g"^2 + "f"^2 - "c"^2)`
= `sqrt(9/4 + 9/4 + 8)`
`= sqrt(18/4 + 8)`
`= sqrt(9/2 + 8)`
`= sqrt((9+16)/2)`
`= sqrt(25/2)`
`= 5/sqrt2`
APPEARS IN
संबंधित प्रश्न
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.
Find the length of the tangent from (1, 2) to the circle x2 + y2 – 2x + 4y + 9 = 0.
The equation of the circle with centre on the x axis and passing through the origin is:
If the centre of the circle is (-a, -b) and radius is `sqrt("a"^2 - "b"^2)` then the equation of circle is:
In the equation of the circle x2 + y2 = 16 then v intercept is (are):
Find the equation of the tangent and normal to the circle x2 + y2 – 6x + 6y – 8 = 0 at (2, 2)
Find centre and radius of the following circles
2x2 + 2y2 – 6x + 4y + 2 = 0
If the equation 3x2 + (3 – p)xy + qy2 – 2px = 8pq represents a circle, find p and q. Also determine the centre and radius of the circle
Choose the correct alternative:
The radius of the circle 3x2 + by2 + 4bx – 6by + b2 = 0 is
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