Question

ab and ac are two chords of the circle having radius R if p and q are distance of the chord ab and AC from the centre respectively and if AB is equal to 2 AC prove that 4 Q square is equal to p square + 3R square.

Answer 1

Given, AB and AC are two chords of a circle of radius r

Again, AB = 2AC and the distances of AB and AC from the center are p and q.

Let AC = x, then AB = 2x

We know that the perpendicular drawn from the centre to the chord bisect the chord.

So, CL = x/2 and AM = x

Now, in ΔOLC,

r² = q² + (x/2)² (By Pythagoras theorem)

=> r² = q² + x² /4

Again, in ΔOAM,

4q² = 4(r² - x² /4) (using 1)

4r² = 4r² - x²

=> 4q² = 4r² - (r² - p² ) (using 2)

=> 4q² = 4r² - r² + p²

=> 4q² = 3r² + p²

Hope this helps!

Answer 2

let ac=a then ab=2a

seg om⊥ac seg on ⊥ seg ab

an=nb=a

am=mc=a

in ΔOMA

OA²=OM²+AM²

     =q²+a²/2²

       =q²+a²/4⇒1

in Δ ONA

OA² =AN²+ON²

      =a²+p² ⇒2

from eq 1 and 2

a²/4 + q²=p²+a²

a²+ 4q²=4p²+4a²

4q²= 4p²+3a²

4q²=p²+3(p²+a²)

4q²=p²+3r²                                                              from eq 2

hence proved