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Question Number 95639 by i jagooll last updated on 26/May/20

$$\mathrm{find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{circle}$$$$\mathrm{containing}\mathrm{the}\mathrm{point}(-2,2)\mathrm{and}$$$$\mathrm{passing}\mathrm{throught}\mathrm{the}\mathrm{points}\mathrm{of}$$$$\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{circle}$$$${\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}-2\mathrm{y}-4=0\mathrm{and}$$$${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-\mathrm{y}-6=0$$

Commented by i jagooll last updated on 26/May/20

$$\mathrm{thank}\mathrm{you}$$

Answered by john santu last updated on 26/May/20

$$\mathrm{substitute}(-2,2)\mathrm{for}(\mathrm{x},\mathrm{y})\mathrm{in}\mathrm{the}$$$$\mathrm{equation}({\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}-2\mathrm{y}-4)+\lambda ({\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-\mathrm{y}-6)=0$$$$\mathrm{then}\lambda =\frac{3}{2}.\mathrm{so}\mathrm{desired}\mathrm{equation}\mathrm{can}\mathrm{be}$$$$\mathrm{written}\mathrm{as}{5\mathrm{x}}^2+{5\mathrm{y}}^2-7\mathrm{y}-26=0$$

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