A locus problem that I enjoyed solving

Geometry Level 4

The locus of poles of tangents to the circle ( x − p ) 2 + y 2 = b 2 (x-p)^2+y^2=b^2 with respect to the circle x 2 + y 2 = a 2 x^2+y^2=a^2 is given by

( a 2 + b x ) 2 = p 2 ( x 2 + y 2 ) (a^2+bx)^2=p^2(x^2+y^2) ( a 2 − b x ) 2 = p 2 ( x 2 + y 2 ) (a^2-bx)^2=p^2(x^2+y^2) ( a 2 + p x ) 2 = b 2 ( x 2 + y 2 ) (a^2+px)^2=b^2(x^2+y^2) ( a 2 − p x ) 2 = b 2 ( x 2 + y 2 ) (a^2-px)^2=b^2(x^2+y^2)

Nihar Mahajan by Nihar Mahajan , 20, India

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