Question

Find the equation of a circle which passes through the points (7,10 and -7,-4 and have radius equal to 10 units

Answer 1

Consider a circle passing through two points A(7,10) and B(-7,-4) and radius 10 units.

Gradient of AB = (10+4)/(7+7) = 1

So gradient of perpendicular bisector of AB = -1.

Midpoint of AB is [(7-7)/2, (10-4)/2], which is (0,3).

Hence, the equation of the perpendicular bisector is y−3=-1(x−0)

The equation can be written as y=-x+3

Let the centre of the circles be (x,y).

Substitute y=-x+3 as the centre lies on the perpendicular bisector.

Then the centre becomes (x,-x+3).

Take the distance between any point on the circle and the centre, which is the radius.

(x−7)^2+(-x+3−10)^2=10^2

(x−7)^2+(-x-7)^2=100

x^2-14x+49+x^2+14x+49=100

2x^2+2×29=100

x^2=21

x=+√21 or x=-√21

y=-√21+3 or y=√21+3

Hence, the possible coordinates of centres are (√21,-√21+3) and (-√21,√21+3).

The equations are

a) (x−√21)^2+(y+√21+3)^2=100 and

b) (x+√21)^2+(y-√21+3)^2=100.

Hope this helps...