Asked by kamau
show that the circle x^2+y^2-6x+4y+2=0 and x^2+y^2+8x+2y-22=0 are orthogonal
Answers
Answered by
Steve
The circles are
(x-3)^2 + (y+2)^2 = 11
(x+4)^2 + (y+1)^2 = 39
So their centers are at (3,-2) and (-4,-1)
The circles intersect at
(73/50 ± √429/50 , -89/50 ± 7√429/50)
or
(1.046,-4.680) and (1.874,1.120)
Now, using those two points, and knowing the centers of the circles, you can show that the lines joining the centers to the intersections are perpendicular.
Since the radii are perpendicular, so are the tangents, and thus the circles.
(x-3)^2 + (y+2)^2 = 11
(x+4)^2 + (y+1)^2 = 39
So their centers are at (3,-2) and (-4,-1)
The circles intersect at
(73/50 ± √429/50 , -89/50 ± 7√429/50)
or
(1.046,-4.680) and (1.874,1.120)
Now, using those two points, and knowing the centers of the circles, you can show that the lines joining the centers to the intersections are perpendicular.
Since the radii are perpendicular, so are the tangents, and thus the circles.
Answered by
ZACHARIA
Solve that question
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