Asked by Anonymous
                Circle C has its center at (9,0) and a point is on the circle at A(8,6).
Which answer verifies whether point P(10,−7) lies on the circle?
a. The statement (9−0)2+(8−6)2≠(9−0)2+(10+7)2 is a true statement, so P is not on ⨀C.
b. The equation (9−0)2+(8−6)2=(9−0)2+(10+7)2 is a true statement, so P is on ⨀C.
c. The statement (9−8)2+(0−6)2≠(9−10)2+(0+7)2 is a true statement, so P is not on ⨀C.
d. The equation (9−8)2+(0−6)2=(9−10)2+(0+7)2 is a true statement, so P is on ⨀C.
            
        Which answer verifies whether point P(10,−7) lies on the circle?
a. The statement (9−0)2+(8−6)2≠(9−0)2+(10+7)2 is a true statement, so P is not on ⨀C.
b. The equation (9−0)2+(8−6)2=(9−0)2+(10+7)2 is a true statement, so P is on ⨀C.
c. The statement (9−8)2+(0−6)2≠(9−10)2+(0+7)2 is a true statement, so P is not on ⨀C.
d. The equation (9−8)2+(0−6)2=(9−10)2+(0+7)2 is a true statement, so P is on ⨀C.
Answers
                    Answered by
            Reiny
            
    This question is a good example of how to make a rather simple concept look complicated
centre is ( 9,0) , so equation is
(x-9)^2 + y^2 = r^2
point (8,6) is on it, then 1^2 + 6^2 = r^2 = 37
the circle is : (x-9)^2 + y^2 = 37
Now sub in P(10,−7)
If it satisfies the equation P is on the circle
If it does not satisfy the equation P is not on the circle
    
centre is ( 9,0) , so equation is
(x-9)^2 + y^2 = r^2
point (8,6) is on it, then 1^2 + 6^2 = r^2 = 37
the circle is : (x-9)^2 + y^2 = 37
Now sub in P(10,−7)
If it satisfies the equation P is on the circle
If it does not satisfy the equation P is not on the circle
                    Answered by
            unknown answer
            
    I think it is on the line is the answer B?
    
                    Answered by
             Anonymous
            
    from where did the 37 came from?
    
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