I appreciate this method of solving the problem but there's another one that makes use of the general equation of circle i.e. x^2 + y^2 + 2gx + 2fy + c = 0 where g , f and c are unknown constants, substitute all the three points in the general equation to get three new equations and solve them to find g, f and c. The values of g, f and c are then substituted in the general equation of the circle to get the desired equation
@@harrystott4707 Certainly solving three simultaneous equations is much quicker (and more universal) method. From general equation of the circle in the form x^2 - 2xY + y^2 - 2yY + C = 0 we get the following three equations: 1) 36 + 12Y + C = 0 2) 16 + 8Y + C = 0, 3) 1 - 2X + 9 + 6Y + C = 0. Subtracting 2) from 1) gives 4Y +20 = 0, Y=-5. Substitution turns 2) into 16-40+C=0, C=24. Substitution of Y and C into 3) gives 1-2x+9-30+24=0, 4-2X=0, X=2. So centre is at (2, -5), and the circle equation is (x-2)^2 + (y+5)^2 - 4 - 25 + 24 = 0, => (x-2)^2 + (y+5)^2 = 5. Therefore R=√5. Easy-peasy.
"If its tricky, we draw a piccy" - I will live my maths career by those words
very clear explanation, so easy to follow along! thanks for the great video.
12:25 rn but never too late for a TL maths video to drop!
Quick question: can you do as level maths stats and mechanics videos for Edexcel?
I appreciate this method of solving the problem but there's another one that makes use of the general equation of circle i.e. x^2 + y^2 + 2gx + 2fy + c = 0 where g , f and c are unknown constants, substitute all the three points in the general equation to get three new equations and solve them to find g, f and c. The values of g, f and c are then substituted in the general equation of the circle to get the desired equation
And which method do you think is quicker💀
@@harrystott4707 Certainly solving three simultaneous equations is much quicker (and more universal) method.
From general equation of the circle in the form x^2 - 2xY + y^2 - 2yY + C = 0 we get the following three equations: 1) 36 + 12Y + C = 0 2) 16 + 8Y + C = 0, 3) 1 - 2X + 9 + 6Y + C = 0. Subtracting 2) from 1) gives 4Y +20 = 0, Y=-5. Substitution turns 2) into 16-40+C=0, C=24. Substitution of Y and C into 3) gives 1-2x+9-30+24=0, 4-2X=0, X=2. So centre is at (2, -5), and the circle equation is (x-2)^2 + (y+5)^2 - 4 - 25 + 24 = 0, => (x-2)^2 + (y+5)^2 = 5. Therefore R=√5. Easy-peasy.
yh yh big up to my mans with the maths
another great vid bro
NICE