# How do you write the equation of the circle whose diameter has endpoints (-1,0) and (6,-5)?

Sep 16, 2016

${x}^{2} + {y}^{2} - 5 x + 5 y - 6 = 0$.

#### Explanation:

If $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ are diametrically opposite pts. of a circle,

then, its eqn. is given by,

$\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) + \left(y - {y}_{1}\right) \left(y - {y}_{2}\right) = 0$.

Applying this, we can immediately get the eqn. of circle as,

$\left(x + 1\right) \left(x - 6\right) + \left(y - 0\right) \left(y + 5\right) = 0$, i.e.,

${x}^{2} + {y}^{2} - 5 x + 5 y - 6 = 0$.

Alternatively,

Knowing that the mid-pt. of a diameter is the Centre of the Circle,

we get the Centre $C \left(\frac{- 1 + 6}{2} , \frac{- 5 + 0}{2}\right) = C \left(\frac{5}{2} , - \frac{5}{2}\right)$.

Now, the pt. $A \left(- 1 , 0\right)$ is on the Circle, and $C \left(\frac{5}{2} , - \frac{5}{2}\right)$ being

the Centre, the dist. $C A$ is the radius $r$ of the Circle. Hence,

$C {A}^{2} = {r}^{2} = {\left(\frac{5}{2} + 1\right)}^{2} = {\left(- \frac{5}{2} - 0\right)}^{2} = \frac{49 + 25}{4} = \frac{74}{4}$

Hence, the eqn of the circle is

(x-5/2)^2+(y+5/2)^2=74/4, or,

x^2+y^2-5x+5y+25/4+25/4-74/4=0, i.e.,

${x}^{2} + {y}^{2} - 5 x + 5 y - 6 = 0$, as before!

Enjoy Maths.!