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Question Number 21925 by >>Umar << last updated on 07/Oct/17

$$\mathrm{In}\mathrm{an}\mathrm{xy}\mathrm{plane}\mathrm{the}\mathrm{graph}\mathrm{of}\mathrm{the}\mathrm{equation}$$$${(\mathrm{x}-6)}^2+{(\mathrm{y}+5)}^2=16\mathrm{is}\mathrm{a}\mathrm{circle}.\mathrm{P}(10,-5)$$$$\mathrm{is}\mathrm{on}\mathrm{the}\mathrm{circle}.\mathrm{If}\mathrm{PQ}\mathrm{is}\mathrm{a}\mathrm{diameter}\mathrm{of}\mathrm{the}$$$$\mathrm{circle},\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{co}-\mathrm{ordinate}\mathrm{at}\mathrm{Q}?$$

Answered by Tinkutara last updated on 07/Oct/17

$$Centerofcircle,C=(6,-5)$$$$LetQ=(x,y)$$$$ThenCisthemidpointofPQ.$$$$\therefore (\frac{10+x}{2},\frac{-5+y}{2})=(6,-5)$$$$10+x=12;y-5=-10$$$$x=2;y=-5$$

Commented by >>Umar << last updated on 07/Oct/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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