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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^2 +y^2 =a^2 and x^2 +y^2 =b^2 respectively are perpendicular.

$${Find}\:{locus}\:{of}\:{point}\:{P}\:{from}\:{which} \\ $$ $${tangents}\:{PA}\:\&\:{PB}\:{to}\:{circles}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$ $${and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \:{respectively}\:{are}\:{perpendicular}. \\ $$

Commented byrahul 19 last updated on 12/Nov/18

Ans→ x^2 +y^2 =a^2 +b^2 .

$${Ans}\rightarrow\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} . \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

tangent on x^2 +y^2 =a^2 from p(α,β) is y−β=m(x−α) y−mx+mα−β=0 distance from (0,0) is radius ∣((0−0+mα−β)/(√(1^2 +m^2 )))∣=a ((mα−β)/(√(1+m^2 )))=a tangent on x^2 +y^2 =b^2 y−β=−(1/m)(x−α) my−mβ+x−α=0 ∣((−mβ−α)/(√(1+m^2 )))∣=b ((mβ+α)/(√(1+m^2 )))=b a^2 +b^2 =(((mα−β)^2 +(mβ+α)^2 )/(1+m^2 )) a^2 +b^2 =((α^2 (1+m^2 )+β^2 (1+m^2 ))/((1+m^2 ))) a^2 +b^2 =α^2 +β^2 hence locus is x^2 +y^2 =a^2 +b^2

$${tangent}\:{on}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:{from}\:{p}\left(\alpha,\beta\right)\:{is} \\ $$ $${y}−\beta={m}\left({x}−\alpha\right) \\ $$ $${y}−{mx}+{m}\alpha−\beta=\mathrm{0} \\ $$ $${distance}\:{from}\:\left(\mathrm{0},\mathrm{0}\right)\:{is}\:{radius} \\ $$ $$\mid\frac{\mathrm{0}−\mathrm{0}+{m}\alpha−\beta}{\sqrt{\mathrm{1}^{\mathrm{2}} +{m}^{\mathrm{2}} }}\mid={a} \\ $$ $$\frac{{m}\alpha−\beta}{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}={a} \\ $$ $${tangent}\:\:{on}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$ $${y}−\beta=−\frac{\mathrm{1}}{{m}}\left({x}−\alpha\right) \\ $$ $${my}−{m}\beta+{x}−\alpha=\mathrm{0} \\ $$ $$\mid\frac{−{m}\beta−\alpha}{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}\mid={b} \\ $$ $$\frac{{m}\beta+\alpha}{\sqrt{\mathrm{1}+{m}^{\mathrm{2}} }}={b} \\ $$ $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\left({m}\alpha−\beta\right)^{\mathrm{2}} +\left({m}\beta+\alpha\right)^{\mathrm{2}} }{\mathrm{1}+{m}^{\mathrm{2}} } \\ $$ $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\frac{\alpha^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)+\beta^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)} \\ $$ $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$ $${hence}\:{locus}\:{is} \\ $$ $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$

Commented byrahul 19 last updated on 13/Nov/18

thanks sir ����

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