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Question Number 214683 by MATHEMATICSAM last updated on 16/Dec/24

Commented by MATHEMATICSAM last updated on 16/Dec/24

$$\mathrm{Circles}\mathrm{C}1\mathrm{and}\mathrm{C}2\mathrm{have}\mathrm{equal}\mathrm{radii}\mathrm{and}$$$$\mathrm{are}\mathrm{tangent}\mathrm{to}\mathrm{the}\mathrm{same}\mathrm{line}\mathrm{XY}.\mathrm{Circle}$$$$\mathrm{C}3\mathrm{is}\mathrm{tangent}\mathrm{to}\mathrm{C}1\mathrm{and}\mathrm{C}2.\mathrm{Find}$$$$\mathrm{distance}h,\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{C}3\mathrm{to}\mathrm{line}$$$$\mathrm{XY}\mathrm{in}\mathrm{terms}\mathrm{of}x\mathrm{and}\mathrm{radii}\mathrm{of}\mathrm{the}\mathrm{circles}.$$$$x\mathrm{is}\mathrm{the}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{centres}\mathrm{of}$$$$\mathrm{C}1\mathrm{and}\mathrm{C}2.$$

Answered by mr W last updated on 16/Dec/24

$$a=radiusofC1andC2$$$$b=radiusofC3$$$${(h-a)}^2+{\left(\frac{x}{2}\right)}^2={(a+b)}^2$$$$\Rightarrow h=a+\sqrt{{(a+b)}^2-\frac{x^2}{4}}$$

Commented by MATHEMATICSAM last updated on 16/Dec/24

$$\mathrm{Thanks}\mathrm{sir}:)$$

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