Find the length of the radius.
[tex]r= \sqrt{(3-(-4))^2+(5-6)^2}
\\r = \sqrt{7^2+(-1)^2}
\\r= \sqrt{50}
\\ r=5 \sqrt{2} [/tex]
Find the length of the diameter.
d = 2r = 2 × 5√2 = 10√2
Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)
[tex]y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
\\y-6= \frac{5-6}{3-(-4)} (x-(-4))
\\y-6=- \frac{1}{7} (x+4)
\\7y-42=-x-4
\\x+7y-38=0[/tex]
The distance from B(x, y) to C(3, 5) is 5√2.
[tex] \sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2}
\\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2
\\(x-3)^2+(y-5)^2=50[/tex]
Solve system of equations.
[tex]x+7y-38=0
\\(x-3)^2+(y-5)^2=50
\\
\\x=38-7y
\\(38-7y-3)^2+(y-5)^2=50
\\(35-7y)^2+(y-5)^2=50
\\1225-490y+49y^2+y^2-10y+25-50=0
\\50y^2-500y+1200=0
\\y^2-10y+24=0
\\y^2-6y-4y+24=0
\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4
[/tex]
[tex]x_1=38-7y_1=38-7 \times 6 = 38-42=-4
\\x_2=38-7y_2=38-7 \times 4 = 38-28=10
[/tex]
Point B could have coordinates
[tex]\\
\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)
[/tex]
But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).
