Dear Student, Please find below the solution to the asked query: We havex2+y2+2ax+c2=0Centre is C1-a,0Radius=r1=a2+0-c2=a2-c2x2+y2+2bx+c2=0Centre is C20,-bRadius=r2=0+b2-c2=b2-c2As circles touch each other, henceC1C2=r1+r2-a-02+0+b2=r1+r2⇒a2+b2=a2-c2+b2-c2Square both sidesa2+b2=a2-c2+b2-c2+2a2-c2b2-c2⇒0=-2c2+2a2-c2b2-c2⇒c2=a2-c2b2-c2Square both sides⇒c4=a2b2-a2c2-b2c2+c4⇒b2c2+a2c2=a2b2Divide each term by a2b2c2⇒1a2+1b2=1c2 Hope this information will clear your doubts about this topic If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible Regards

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Please find below the solution to the asked query:

We have x^{2} + y^{2} + 2 ax + c^{2} = 0 Centre is C_{1} (- a, 0) Radius = r_{1} = \sqrt{a^{2} + 0 - c^{2}} = \sqrt{a^{2} - c^{2}} x^{2} + y^{2} + 2 bx + c^{2} = 0 Centre is C_{2} (0, - b) Radius = r_{2} = \sqrt{0 + b^{2} - c^{2}} = \sqrt{b^{2} - c^{2}} As circles touch each other, hence C_{1} C_{2} = r_{1} + r_{2} \sqrt{{(- a - 0)}^{2} + {(0 + b)}^{2}} = r_{1} + r_{2} \Rightarrow \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} - c^{2}} + \sqrt{b^{2} - c^{2}} Square both sides a^{2} + b^{2} = a^{2} - c^{2} + b^{2} - c^{2} + 2 \sqrt{a^{2} - c^{2}} \sqrt{b^{2} - c^{2}} \Rightarrow 0 = - 2 c^{2} + 2 \sqrt{a^{2} - c^{2}} \sqrt{b^{2} - c^{2}} \Rightarrow c^{2} = \sqrt{a^{2} - c^{2}} \sqrt{b^{2} - c^{2}} Square both sides \Rightarrow c^{4} = a^{2} b^{2} - a^{2} c^{2} - b^{2} c^{2} + c^{4} \Rightarrow b^{2} c^{2} + a^{2} c^{2} = a^{2} b^{2} Divide each term by a^{2} b^{2} c^{2} \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}

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