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Question Number 140597 by bramlexs22 last updated on 10/May/21

$$\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{which}$$$$\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}(2,0)\mathrm{and}$$$$\mathrm{whose}\mathrm{center}\mathrm{is}\mathrm{the}\mathrm{limit}\mathrm{of}\mathrm{the}$$$$\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}$$$$\mathrm{lines}3\mathrm{x}+5\mathrm{y}=1\mathrm{and}(2+\mathrm{c})\mathrm{x}+{5\mathrm{c}}^2\mathrm{y}=1$$$$\mathrm{as}\mathrm{c}\to 1.$$

Answered by TheSupreme last updated on 10/May/21

$$3x+5y=1$$$$(2+c)x+5c^{2}y=1$$$$(c-1)x+5(c^2-1)y=0$$$$y=\frac{x}{5(c+1)}$$$$3x+\frac{x}{c+1}=1\to x\left(\frac{4c+3}{c+1}\right)=1$$$$x=\frac{c+1}{4c+3}\to \frac{2}{7}$$$$y\to \frac{1}{35}$$$${(x-\frac{2}{7})}^2+{(y-\frac{1}{35})}^2=R^2$$$$(2,0)$$$${\left(\frac{12}{7}\right)}^2+{\left(\frac{1}{35}\right)}^2=R^2$$$$\frac{1}{7^2}(144+\frac{1}{25})=R^2$$$$\frac{361}{{35}^2}=R^2$$$${(x-\frac{2}{7})}^2+{(y-\frac{1}{35})}^2=\frac{361}{{35}^2}$$

Commented by bramlexs22 last updated on 10/May/21

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

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