Given that 2y - x = 5 is a tangent to the circle with centre (1,-2), find the equation of the circle. Answer:x^2 + y^2 - 2x + 4y - 15 =0.
In urgent need of help. Trying to solve this question too. Hopefully will get a reply by 10pm? Thanks all!
Originally posted by anpanman:Given that 2y - x = 5 is a tangent to the circle with centre (1,-2), find the equation of the circle. Answer:x^2 + y^2 - 2x + 4y - 15 =0.
In urgent need of help. Trying to solve this question too. Hopefully will get a reply by 10pm? Thanks all!
Equation for the circle : (x-1)^2 + (y+2)^2 = r^2, where r = radius
Equation for line: 2y - x = 5 => x = 2y - 5
Sub the line equation into the circle equation.
You will get 5y^2 - 20y + ( 40 - r^2 ) = 0
Here comes the part where it might be a bit hard to explain to you.
If you reshuffle the new equation, you should get ( 5y - a )( y - b) = 0, where a and b are solutions. Then you will realise one thing, since it is a tangent to the circle , there should only be one answer.
Hence, 5y - a = 0 and y - b = 0 should fetch you the same value for y.
= > a = 5b, and the equation can be reform into ( 5y - 5b )( y - b) = 0
With this new understanding, use the cross divison method on the equation 5y^2 - 20y + ( 40 - r^2 ) = 0
You will see that -5by + ( -5by ) = -20 y, and hence b = 2
You will also realise that -5b X -b = ( 40 - r^2 ), and by subbing b = 2 in, you will get r^2 =20
Hence, you will get your equation (x-1)^2 + (y+2)^2 = 20, which simplfies to x^2 + y^2 -2x + 4y -15 = 0