Answer:
(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
