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Question Number 123393 by malwan last updated on 25/Nov/20

$$findtheequationsof\left[two\right]circles$$$$whichthiercenterboth(2,-2)$$$$andtangentwithcircle$$$$x^2+y^2-8x+10y+5=0$$

Commented by malwan last updated on 25/Nov/20

$$canyoudrowitsir?$$

Commented by MJS_new last updated on 25/Nov/20

$$\mathrm{sorry}\mathrm{I}\mathrm{have}\mathrm{no}\mathrm{drawing}\mathrm{app}$$

Commented by talminator2856791 last updated on 25/Nov/20

$$\mathrm{you}\mathrm{can}\mathrm{use}\mathrm{this}\mathrm{app}\mathrm{drawing}\mathrm{tool}$$

Commented by malwan last updated on 25/Nov/20

$$canyoutrysir?$$

Commented by MJS_new last updated on 25/Nov/20

$$\mathrm{sorry}\mathrm{I}\mathrm{never}\mathrm{used}\mathrm{it}\mathrm{and}\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{the}\mathrm{time}$$$$\mathrm{to}\mathrm{get}\mathrm{used}\mathrm{to}\mathrm{it}.\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{sketch}$$

Commented by talminator2856791 last updated on 25/Nov/20

$$\mathrm{yes}\mathrm{i}\mathrm{tried}\mathrm{but}\mathrm{it}\mathrm{is}\mathrm{difficult}\mathrm{to}\mathrm{use}\mathrm{this}$$

Answered by talminator2856791 last updated on 25/Nov/20

$$$$$$x^2+y^2-8x+10y+5={(x-4)}^2+{(y+5)}^2+5-16-25$$$${(x-4)}^2+{(y+5)}^2=r^2=36$$$$r=6$$$$r(4;-5)$$$$\mathrm{distance}\mathrm{between}\mathrm{centres}d$$$$d=\sqrt{2^2+3^2}$$$$d=\sqrt{13}$$$$\sqrt{13}\ll r$$$$$$$$\mathrm{equations}\mathrm{of}\mathrm{circles}:$$$${(x-2)}^2+{(y+2)}^2={(6-\sqrt{13})}^2$$$${(x-2)}^2+{(y+2)}^2={(6+\sqrt{13})}^2$$

Commented by malwan last updated on 25/Nov/20

$$thankyousir$$

Answered by MJS_new last updated on 25/Nov/20

$$\left(1\right){(x-2)}^2+{(y+2)}^2-r^2=0$$$$\left(2\right)x^2+y^2-8x+10y+5=0$$$$\left(1\right)x^2+y^2-4x+4y+8-r^2=0$$$$\left(1\right)-\left(2\right)4x-6y+3-r^2=0\Rightarrow y=\frac{4x-r^2+3}{6}$$$$\mathrm{insert}\mathrm{in}\left(1\right)\mathrm{or}\left(2\right)$$$$\frac{13}{9}{x}^2-\frac{2(r^2+3)}{9}+\frac{r^4-66r^2+369}{36}=0$$$$x^2-\frac{2(o^2+3)}{13}x+\frac{r^4-66r^2+369}{52}=0$$$$\mathrm{tangenting}\mathrm{circles}\mathrm{means}\mathrm{we}\mathrm{must}\mathrm{find}r\mathrm{in}$$$$\mathrm{order}\mathrm{to}\mathrm{get}\mathrm{a}?\mathrm{double}\mathrm{solution}\mathrm{for}x$$$$x^2+px+q=0\Rightarrow x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-q}$$$$\mathrm{we}\mathrm{need}\frac{p^2}{4}-q=0$$$$\Rightarrow \frac{9}{676}(r^4-98r^2+529)=0$$$$\Rightarrow r=6\pm \sqrt{13}[r>0]$$$$\Rightarrow$$$$\mathrm{circle}1:{(x-2)}^2+{(y+2)}^2+49-12\sqrt{13}=0$$$$\mathrm{circle}2:{(x-2)}^2+{(y+2)}^2+49+12\sqrt{13}=0$$

Commented by malwan last updated on 25/Nov/20

$$thankyousomuchsir$$

Commented by peter frank last updated on 25/Nov/20

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

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