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 | ||
$${C}_{\mathrm{1}} \:\:\left({x}−{a}\right)^{\mathrm{2}} +\left({y}−{b}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \:\:\:\left[{r}=\frac{{d}}{\mathrm{2}}=\frac{\mathrm{6}}{\mathrm{2}}=\mathrm{3}\right] \\ $$$$\mid\frac{\mathrm{5}{a}+\mathrm{12}{b}−\mathrm{10}}{\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}\mid=\mid\frac{\mathrm{5}{a}−\mathrm{12}{b}−\mathrm{40}}{\sqrt{\mathrm{5}^{\mathrm{2}} +\left(−\mathrm{12}\right)^{\mathrm{2}} }}\mid=\mathrm{3} \\ $$$$\mathrm{5}{a}+\mathrm{12}{b}−\mathrm{10}=\mathrm{39} \\ $$$$\mathrm{5}{a}−\mathrm{12}{b}−\mathrm{40}=\mathrm{39} \\ $$$$\mathrm{10}{a}=\mathrm{78}+\mathrm{50}\rightarrow{a}=\mathrm{12}.\mathrm{8} \\ $$$$\mathrm{24}{b}+\mathrm{30}=\mathrm{0}\rightarrow{b}=−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${C}_{\mathrm{1}} \:\left({x}−\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} \\ $$$${C}_{\mathrm{2}} \:\:\left({x}−\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{R}^{\mathrm{2}} \\ $$$${distance}\:{from}\:\left(\mathrm{12}.\mathrm{8}\:,\frac{\mathrm{5}}{\mathrm{4}}\right)\:{to}\:{line}\:\mathrm{5}{x}+\mathrm{12}{y}−\mathrm{10}=\mathrm{0} \\ $$$$\mid\frac{\mathrm{5}×\mathrm{12}.\mathrm{8}−\mathrm{12}×\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{10}}{\sqrt{\mathrm{5}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}\mid={h}\left({say}\right) \\ $$$$ \\ $$$$\mid\frac{\mathrm{39}}{\mathrm{13}}\mid={h} \\ $$$${R}^{\mathrm{2}} ={h}^{\mathrm{2}} +\left(\frac{\mathrm{8}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} =\mathrm{9}+\mathrm{16}=\mathrm{25} \\ $$$${C}_{\mathrm{2}} \:\:\:\left({x}−\mathrm{12}.\mathrm{8}\right)^{\mathrm{2}} +\left({y}+\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{25} \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by ajfour last updated on 24/Jun/19 | ||
$${please}\:{check},\:{Sir}.. \\ $$ | ||