m4 Let Gamma1 be a circle of center C1 and radius r and Gamm

m4

Let Gamma_1 be a circle of center C_1 and radius r, and Gamma_2 another circle which intersects Gamma_1 in the points A and B. In class we have shown that Gamma_1 is perpendicular to Gamma_2 then I_C, r(Gamma_2) = Gamma_2. Here you are requested to show the converse implication: if I_c1, r(Gamma_2) = Gamma_2 then Gamma_1 is perpendicular to Gamma_2.

Solution

let radius of circle with centre C₂ is R

and let x = length of DC₁.

now C₁D *C₁E = r^2

C₁D = x

C₁E = x + 2R

so x*(x+2R) = r^2

adding R^2 on both sides

we get

R^2+ x*(x+2R) = r^2 +R^2

or (R+x)^2 = r^2 + R^2

now in traingle C₁AC₂

AC₁ = r

AC₂ = R

C₁C₂ = R+x

since

(R+x)^2 = r^2 + R^2 , it is right angled triangle at A.

hence proved