Find the equation of the line of the circle passing through the points (1,2) (3,-4) (5,-6)

(x-11)² + (y-2)² = 10²

Step-by-step explanation:

Given circle passes through

P(1,2), Q(3,-4), R(5,-6)

Assume following circle with centre O(a,b) and radius r contains P,Q,R.

(x-a)²+(y-b)² = r^2 (1)

Substitute P in (1) and expand, simplify

(1-a)²+(2-b)²=r²

a²+b² -2a-4b+5-r²=0  (2)

Similarly, substitute Q in (1) and expand, simplify

(3-a)²+(-4-b)² = r²

a²+b²-6a+8b+25-r²=0  (3)

Similarly, substitute R in (1) and expand, simplify

(5-a)²+(-6-b)² = r²

a²+b²-10a+12b+61-r²=0  (4)

We can now eliminate a^2 and b^2 by simple subtractions

(2) - (3)

4a-12b-20 = 0   (5)

(3) - (4)

4a-4b-36 = 0   (6)

Finally, (6) - (5) and solve for b

8b - 16 = 0

b=2   (7)

substitute b in (5) and solve for a

4a-12(2) -20 =0

4a=44

a=11  (8)

Finally, substitute a & b in (2) to solve for r

r^2 = 11²+2²-2*11-4*2+5 = 121+4-22-8+5 = 100

r=sqrt(100) = 10  (positive value only)

therefore the equation of the circle is

(x-11)² + (y-2)² = 10²