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Prove that the circles x²+²-4x+6y + 8 = 0 and x² + y² - 10x - 6y + 14 = 0 touch at the point (3,-1).
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Need a written solution.....
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Prove that the circles x²+²-4x+6y + 8 = 0 and x² + y² - 10x - 6y + 14 = 0 touch at the point (3,-1).
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Answers & Comments
Verified answer
Apply the following geometric principles: (0) Circles touch if d-r, r, where d is the distance between the centres and r. r, the radi of the circles, (i) The point of contact divides the central line in the ratio of their radii internally for external
contact and externally for internal contact. Here, CC2-3) and r,-√4-9-5-√5C, (5, 3) andr√25-9-14-25
Hence, the given circles touch externally Now, the point of contact (x, y) divides C, C, in the ratio 1:2.
<-3
142
my + my
2x(-3)+1x3
Answer:
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