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GeometryQuestion and Answers: Page 49

Question Number 174283    Answers: 1   Comments: 1

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Question Number 174185    Answers: 1   Comments: 0

The circle x^2 +y^2 −2x−4y−20=0 is inscribed in a square. One vertex of the square is (−4, 7). Find the coordinates of the other vertices.

$$\mathrm{The}\:\mathrm{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{4}{y}−\mathrm{20}=\mathrm{0}\:\mathrm{is} \\ $$$$\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{square}.\:\mathrm{One}\:\mathrm{vertex} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{is}\:\left(−\mathrm{4},\:\mathrm{7}\right).\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{coordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{vertices}. \\ $$

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The circles x^2 +y^2 −2ax+8y+13=0 and x^2 +y^2 +2x+2by+1=0 are congruent. If they are 2(√(10)) units apart, find the possible values of a and b.

$$\mathrm{The}\:\mathrm{circles}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ax}+\mathrm{8}{y}+\mathrm{13}=\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}{by}+\mathrm{1}=\mathrm{0}\:\mathrm{are}\: \\ $$$$\mathrm{congruent}.\:\mathrm{If}\:\mathrm{they}\:\mathrm{are}\:\mathrm{2}\sqrt{\mathrm{10}}\:\mathrm{units}\: \\ $$$$\mathrm{apart},\:\mathrm{find}\:\mathrm{the}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of} \\ $$$${a}\:\mathrm{and}\:{b}. \\ $$

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