Question

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

Answer 1

`x^2+y^2-5x+5y-6=0`.

Explanation:

If `(x_1,y_1) and (x_2,y_2)` are diametrically opposite pts. of a circle,

then, its eqn. is given by,

`(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0`.

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

`(x+1)(x-6)+(y-0)(y+5)=0`, i.e.,

`x^2+y^2-5x+5y-6=0`.

Alternatively,

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

we get the Centre `C((-1+6)/2,(-5+0)/2)=C(5/2,-5/2)`.

Now, the pt. `A(-1,0)` is on the Circle, and `C(5/2,-5/2)` being

the Centre, the dist. `CA` is the radius `r` of the Circle. Hence,

`CA^2=r^2=(5/2+1)^2=(-5/2-0)^2=(49+25)/4=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^2-5x+5y-6=0`, as before!

Enjoy Maths.!