Question

AB and AC are two chord of a circle of radius r, such that AB=AC . if p and q are the distances AB and AC from the centre . prove that 4q2=p2+3r2

Answer 1

Hi there !

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Given :

AB and AC are two chords of a circle of radius r such that AB= 2AC.

P and q are the perpendicular distances of AB and AC from the centre O, that is

OM = p and ON = q

And 'r' is the radius of the circle.

To prove :

$\mathsf {4q {}^{2} = p {}^{2} + 3r {}^{2}}$

Construction ; Join OA.

PROOF :

OM and ON are the perpendicular distances of AB and AC from the centre O.

AN = $\mathsf {\frac {AC} {2} }$

AM = $\mathsf {\frac {AB} {2} }$

[Because these are perpendicular from center to chord]

In right angled ΔONA,

$\mathsf{ON {} ^{2} + NA{} ^{2} = OA{} ^{2}}$

⇒ $\mathsf{q {} ^{2} + NA{} ^{2} = r{} ^{2}}$

⇒ $\mathsf{NA {} ^{2} = r{} ^{2} - q{} ^{2}} .... (i)$

In right angled ΔONA,

$\mathsf{OM {} ^{2} + AM{} ^{2} = OA{} ^{2}}$

⇒ $\mathsf{p {} ^{2} + AM{} ^{2} = r{} ^{2}}$

⇒ $\mathsf{AM {} ^{2} = r{} ^{2} - p{} ^{2}}....(ii)$

Since,

AM = $\mathsf {\frac {AB} {2} =\frac {2AC} {2} = AC = 2NA}$

Now, from (i) and (ii), we have ;

$\mathsf{r {}^{2} - p {}^{2} = AM {}^{2} = (2NA) {}^{2} = 4NA {}^{2} } \\ \\ \mathsf{r {}^{2} - p {}^{2} = 4[r {}^{2} - q {}^{2} ] } \\ \\ \mathsf{r {}^{2} - p {}^{2} = 4r {}^{2} - 4q {}^{2} } \\ \\ \mathsf{4 q {}^{2} = 3r {}^{2} + p {}^{2} }$

Hence, it is proved !

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Thanks for the question !

☺️❤️☺️

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