Question Number 62698 by Ankit0512 last updated on 24/Jun/19
$${Lines}\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{10}=\mathrm{0}\:{and}\:\mathrm{5}{x}−\mathrm{12}{y}−\mathrm{40}=\mathrm{0} \\ $$$${touch}\:{circle}\:{C}_{\mathrm{1}} \:{of}\:{diameter}\:\mathrm{6}.\:{If}\:{the}\: \\ $$$${center}\:{of}\:{C}_{\mathrm{1}} \:{lies}\:{in}\:{the}\:{Ist}\:{quadrant}, \\ $$$${find}\:{the}\:{equation}\:{of}\:{circle}\:{C}_{\mathrm{2}} \:{which}\:{is}\: \\ $$$${concentric}\:{with}\:{C}_{\mathrm{1}\:} \:{and}\:{cuts}\:{intercept} \\ $$$${of}\:{length}\:\mathrm{8}\:{on}\:{these}\:{lines}. \\ $$
Answered by tanmay last updated on 24/Jun/19
![C_1 (x−a)^2 +(y−b)^2 =3^2 [r=(d/2)=(6/2)=3] ∣((5a+12b−10)/(√(5^2 +12^2 )))∣=∣((5a−12b−40)/(√(5^2 +(−12)^2 )))∣=3 5a+12b−10=39 5a−12b−40=39 10a=78+50→a=12.8 24b+30=0→b=−(5/4) C_1 (x−12.8)^2 +(y+(5/4))^2 =3^2 C_2 (x−12.8)^2 +(y+(5/4))^2 =R^2 distance from (12.8 ,(5/4)) to line 5x+12y−10=0 ∣((5×12.8−12×(5/4)−10)/(√(5^2 +12^2 )))∣=h(say) ∣((39)/(13))∣=h R^2 =h^2 +((8/2))^2 R^2 =9+16=25 C_2 (x−12.8)^2 +(y+(5/4))^2 =25](Q62701.png)
$${C}_{\mathrm{1}} \:\:\left({x}−{a}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \:\:\:\left[{r}=\frac{{d}}{\mathrm{2}}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3}\right] \\ $$$$\mid\frac{\mathrm{5}{a}+\mathrm{12}{b}−\mathrm{10}}{\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}\mid=\mid\frac{\mathrm{5}{a}−\mathrm{12}{b}−\mathrm{40}}{\sqrt{\mathrm{5}^{\mathrm{2}} +\left(−\mathrm{12}\right)^{\mathrm{2}} }}\mid=\mathrm{3} \\ $$$$\mathrm{5}{a}+\mathrm{12}{b}−\mathrm{10}=\mathrm{39} \\ $$$$\mathrm{5}{a}−\mathrm{12}{b}−\mathrm{40}=\mathrm{39} \\ $$$$\mathrm{10}{a}=\mathrm{78}+\mathrm{50}\rightarrow{a}=\mathrm{12}.\mathrm{8} \\ $$$$\mathrm{24}{b}+\mathrm{30}=\mathrm{0}\rightarrow{b}=−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${C}_{\mathrm{1}} \:\left({x}−\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \\ $$$${C}_{\mathrm{2}} \:\:\left({x}−\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} \\ $$$${distance}\:{from}\:\left(\mathrm{12}.\mathrm{8}\:,\frac{\mathrm{5}}{\mathrm{4}}\right)\:{to}\:{line}\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{10}=\mathrm{0} \\ $$$$\mid\frac{\mathrm{5}×\mathrm{12}.\mathrm{8}−\mathrm{12}×\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{10}}{\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}\mid={h}\left({say}\right) \\ $$$$ \\ $$$$\mid\frac{\mathrm{39}}{\mathrm{13}}\mid={h} \\ $$$${R}^{\mathrm{2}} ={h}^{\mathrm{2}} +\left(\frac{\mathrm{8}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} =\mathrm{9}+\mathrm{16}=\mathrm{25} \\ $$$${C}_{\mathrm{2}} \:\:\:\left({x}−\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$$ \\ $$$$ \\ $$
Commented by ajfour last updated on 24/Jun/19
$${please}\:{check},\:{Sir}.. \\ $$