Thread: Equation of Circle from right-angled triangle

1. Equation of Circle from right-angled triangle

The point D, E and F have coordinates (-2,0), (0,-1) and (2,3) respectively.

i) gradient of DE = -0.5
ii) 0.5x+y-4=0 line F parallel to DE

iii) by calculating the gradient of EF, show that DEF is a right-angled triangle.
iv) calculate the length of DF.
v) Use the results of parts(iii) and (iv to show that the circle which passes through D,E and F has equation $x^2+y^2-3y-4=0.$

i've done i),ii),iii),iv)

i) gradient of DE = -0.5
ii) 0.5x+y-4=0 line F parallel to DE
iii) y= 2x-1 = grad of normal
which bisect DE at some point therefore it's a right angled triangle.
iv) $sqroot(4^2+3^2)$

thank you for helping!

2. Do you have to use the earlier parts? If so, are you allowed to use the result (from geometry) relating right angles and triangles?

If so, then find the midpoint of the hypotenuse, as this will be the center of the circle. You've found the length of the hypotenuse; now divide by two to find the radius value.

Once you have the center and the radius, you can plug into the center-radius form of the circle equation, multiply everything out, and rearrange to get the required form of the equation.

3. Originally Posted by stapel
If so, then find the midpoint of the hypotenuse, as this will be the center of the circle. You've found the length of the hypotenuse; now divide by two to find the radius value.
ok i understand this part which leave me with 2.5 as the radius.

Once you have the center and the radius, you can plug into the center-radius form of the circle equation, multiply everything out, and rearrange to get the required form of the equation.
but how do i find out the centre points of the circle?