The centre will be at the middle of these points so it is the mean value.
Let point 1 be P_1->(x_1,y_1)->(-3,4)P1→(x1,y1)→(−3,4)
Let point 2 be P_2->(x_2,y_2)->(4,-3)P2→(x2,y2)→(4,−3)
Let the centre be P_c->(x_x,y_c)Pc→(xx,yc)
Let the radius be rr
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color(blue)("Determine the centre")Determine the centre
Mean for x-> (4+(-3))/2= 1/2 ->x_c→4+(−3)2=12→xc
Mean for y->(4+(-3))/2=1/2->y_c→4+(−3)2=12→yc
color(blue)("Centre "-> (x_c,y_c)->(1/2,1/2))Centre →(xc,yc)→(12,12)
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color(blue)("Determine the magnitude of the radius")Determine the magnitude of the radius
Using Pythagoras
I chose (x_1,y_1)(x1,y1) to determine the length of rr
r^2=(x_c-x_1)^2+(y_c-y_1)^2r2=(xc−x1)2+(yc−y1)2
r^2=[1/2-(-3)]^2+[1/2-4]^2r2=[12−(−3)]2+[12−4]2
r=sqrt(49/4+49/4) = sqrt(49/2)r=√494+494=√492
r=7/sqrt(2)r=7√2
Multiply by 1 but in the form of 1=sqrt(2)/sqrt(2)1=√2√2
color(blue)(r=(7sqrt(2))/2" "->" " r^2=49/2)r=7√22 → r2=492
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color(blue)("Putting it all together")Putting it all together
Let any xx be x_ixi
Let any yy be y_iyi
The equation of this circle is
r^2=(x_i-x_x)^2+(y_i-y_c)^2r2=(xi−xx)2+(yi−yc)2
color(blue)(=>49/2=(x_i-1/2)^2+(y_i-1/2)^2)⇒492=(xi−12)2+(yi−12)2
