Question

Convert the general form of the circle given into standard form.

Answer 1

$The\ standard\ form\ of\ the\ circle:\\\\(x-a)^2+(y-b)^2=r^2\\\\where\\(a;\ b)\ are\ the\ coordinates\ of\ a\ center\ of\ the\ circle\\r\ is\ a\ radius\ of\ the\ circle\\------------------------$

$2x^2+2y^2-20x-8y+50=0\ \ \ \ |divide\ both\ sides\ by\ 2\\\\x^2+y^2-10x-4y+25=0\\\\x^2-10x+y^2-4y=-25\\\\x^2-2x\cdot5+y^2-2y\cdot2=-25\\\\\underbrace{x^2-2x\cdot5+5^2}_{(*)}-5^2+\underbrace{y^2-2y\cdot2+2^2}_{(*)}-2^2=-25\\(x-5)^2-25+(y-2)^2-4=-25\\\\(x-5)^2+(y-2)^2-29=-25\\\\(x-5)^2+(y-2)^2=-25+29$

$\boxed{(x-5)^2+(y-2)^2=4}\\\\the\ center:(5;\ 2)\\the\ radius:r=\sqrt4=2$



$(*)\ (a-b)^2=a^2-2ab+b^2$

Answer 2

$"Standard Form" \ for \ the \ equation \ of \ a \ circle : \\\\(x-a)^2+(y-b)^2=r^2 \\ the \ center \ (a,b)\\radius \ r \\\\ 2x^2 + 2y^2 -20x - 8y + 50 = 0 \ \ \ Divide \ thru \ by \ 2 \\\\x^2 + y^2 -10x - 4y + 25 = 0$

$x^2-10x +25 +y^2-4y +4-4 =0 \\\\(x^2-10x +25) +(y^2-4y +4)=4 \\\\(x-5)^2+(y-2)^2=2^2$