Question

Two circles with centre O and P intersect each others in point C and D. Chord AB of the circle with centre O touches the circle with centre P in point E
Prove that

Answer 1

Let Q is perpendicular drop of P on OO'.

Let the radii of the circles are R and r.

Let the spacing (perpendicular distance) between OO' and CD is d.

OQ=√(R^2-d^2)

QO'=√(r^2-d^2)

OO'=[√(R^2-d^2)+√(r^2-d^2)]=x (say).

CP=2√(R^2-d^2)+2√(r^2-d^2)

=2[(√R^2-d^2)+√(r^2-d^2)=2x=2.OO'.

(Proved).

Answer 2

Answer:

OQ=√(R^2-d^2)

QO'=√(r^2-d^2)

OO'=[√(R^2-d^2)+√(r^2-d^2)]=x (say).

CP=2√(R^2-d^2)+2√(r^2-d^2)

=2[(√R^2-d^2)+√(r^2-d^2)=2x=2.OO'.

Step-by-step explanation: