π΄(1,β2) is a point on the circle (π₯β3)2+ (π¦+1)2= 5
a. State the coordinates of the centre of the circle and hence find the coordinates of the point π΅ where π΄π΅ is the diameter of the circle.
b. πΆ(2,1) also lies on the circle. Use coordinate geometry to verify that angle π΄πΆπ΅ = 900
a.
The coordinates of the centre of the circle are "D(3, -1)."
If "AB" is the diameter of the circle, then "D" is the midpoint of "AB"
"y_D=\\dfrac{y_A+y_B}{2}=>y_B=2y_D-y_A"
"x_B=2(3)-1=5, y_B=2(-1)-(-2)=0"
Point "B(5, 0)"
b.
Point "C(2, 1)"
"\\overrightarrow{CB}=\\langle5-2, 0-1\\rangle"
"\\overrightarrow{CA}\\cdot \\overrightarrow{CB}=-1(3)+(-3)(-1)=-3+3=0"
"|\\overrightarrow{CA}|\\not=0, |\\overrightarrow{CB}|\\not=0"
Therefore "\\overrightarrow{CA}\\perp \\overrightarrow{CB}," and "\\angle ACB=90\\degree."
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