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

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Answer 1 · 2016-09-16T18:03:12

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

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.!