Lines-5x-12y-10-0-and-5x-12y-40-0-touch-circle-C-1-of-diameter-6-If-the-center-of-C-1-lies-in-the-Ist-quadrant-find-the-equation-of-circle-C-2-which-is-concentric-with-C-1-and-cuts-intercep

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Question Number 62698 by Ankit0512 last updated on 24/Jun/19

$$Lines5x+12y-10=0and5x-12y-40=0$$$$touchcircle{C}_{1}ofdiameter6.Ifthe$$$$centerof{C}_{1}liesintheIstquadrant,$$$$findtheequationofcircle{C}_{2}whichis$$$$concentricwith{C}_{1}andcutsintercept$$$$oflength8ontheselines.$$

Answered by tanmay last updated on 24/Jun/19

$$C_1{(x-a)}^2+{(y-b)}^2=3^2[r=\frac{d}{2}=\frac{6}{2}=3]$$$$\mid \frac{5a+12b-10}{\sqrt{5^2+{12}^2}}\mid =\mid \frac{5a-12b-40}{\sqrt{5^2+{(-12)}^2}}\mid =3$$$$5a+12b-10=39$$$$5a-12b-40=39$$$$10a=78+50\to a=12.8$$$$24b+30=0\to b=-\frac{5}{4}$$$$C_1{(x-12.8)}^2+{(y+\frac{5}{4})}^2=3^2$$$$C_2{(x-12.8)}^2+{(y+\frac{5}{4})}^2={\mathrm{R}}^2$$$$distancefrom(12.8,\frac{5}{4})toline5x+12y-10=0$$$$\mid \frac{5\times 12.8-12\times \frac{5}{4}-10}{\sqrt{5^2+{12}^2}}\mid =h\left(say\right)$$$$$$$$\mid \frac{39}{13}\mid =h$$$$R^2=h^2+{\left(\frac{8}{2}\right)}^2$$$$R^2=9+16=25$$$$C_2{(x-12.8)}^2+{(y+\frac{5}{4})}^2=25$$$$$$$$$$

Commented by ajfour last updated on 24/Jun/19

$$pleasecheck,Sir..$$

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