Advertisements
Advertisements
प्रश्न
Find the centre and radius of the circle.
5x2 + 5y2+ 4x – 8y – 16 = 0
उत्तर
To make coefficient of x2 unity, divide the equation by 5 we get,
`x^2 + y^2 + 4/5 x - 8/5 y - 16/5 = 0`
Comparing the above equation with x2 + y2 + 2gx + 2fy + c = 0 we get,
2g = `4/5`, 2f = `-8/5`, c = `(- 16)/5`
∴ g = `2/5`, f = `- 4/5`, c = `(-16)/5`
Centre = (-g, -f) = `((-2)/5, 4/5)`
Radius = `sqrt("g"^2 + "f"^2 - "c"^2)`
`= sqrt((2/5)^2 + ((-4)/5)^2 - ((-16)/5))`
`= sqrt(4/25 + 16/25 + 16/5)`
`= sqrt((4 + 16 + 16 xx 5)/25)`
`= sqrt((20 + 80)/25)`
`= sqrt(100/25)`
`= sqrt4` = 2
APPEARS IN
संबंधित प्रश्न
Find the equation of the circle whose centre is (-3, -2) and having circumference 16π.
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 (4, 1) is one extremity of a diameter of the circle x2 + y2 - 2x + 6y - 15 = 0 find the other extremity.
The centre of the circle x2 + y2 – 2x + 2y – 9 = 0 is:
The equation of the circle with centre (3, -4) and touches the x-axis is:
If the circle touches the x-axis, y-axis, and the line x = 6 then the length of the diameter of the circle is:
Obtain the equation of the circle for which (3, 4) and (2, -7) are the ends of a diameter.
A circle of area 9π square units has two of its diameters along the lines x + y = 5 and x – y = 1. Find the equation of the circle
Choose the correct alternative:
The equation of the circle passing through (1, 5) and (4, 1) and touching y-axis `x^2 + y^2 - 5x - 6y + 9 + lambda(4x + 3y - 19)` = where `lambda` is equal to
Choose the correct alternative:
The equation of the normal to the circle x2 + y2 – 2x – 2y + 1 = 0 which is parallel to the line 2x + 4y = 3 is
