Question Number 47621 by rahul 19 last updated on 12/Nov/18
$${Find}\:{locus}\:{of}\:{point}\:{P}\:{from}\:{which} \\ $$$${tangents}\:{PA}\:\&\:{PB}\:{to}\:{circles}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \:{respectively}\:{are}\:{perpendicular}. \\ $$
Commented by rahul 19 last updated on 12/Nov/18
$${Ans}\rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} . \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
$${tangent}\:{on}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:{from}\:{p}\left(\alpha,\beta\right)\:{is} \\ $$$${y}−\beta={m}\left({x}−\alpha\right) \\ $$$${y}−{mx}+{m}\alpha−\beta=\mathrm{0} \\ $$$${distance}\:{from}\:\left(\mathrm{0},\mathrm{0}\right)\:{is}\:{radius} \\ $$$$\mid\frac{\mathrm{0}−\mathrm{0}+{m}\alpha−\beta}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +{m}^{\mathrm{2}} }}\mid={a} \\ $$$$\frac{{m}\alpha−\beta}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}={a} \\ $$$${tangent}\:\:{on}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$${y}−\beta=−\frac{\mathrm{1}}{{m}}\left({x}−\alpha\right) \\ $$$${my}−{m}\beta+{x}−\alpha=\mathrm{0} \\ $$$$\mid\frac{−{m}\beta−\alpha}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\mid={b} \\ $$$$\frac{{m}\beta+\alpha}{\:\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}={b} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\left({m}\alpha−\beta\right)^{\mathrm{2}} +\left({m}\beta+\alpha\right)^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\alpha^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)+\beta^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$$${hence}\:{locus}\:{is} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$
Commented by rahul 19 last updated on 13/Nov/18
thanks sir ����