Question

Find the valuue of m so that the quadratic equation mx(x-7)49=0 hastwo equal rootpa and pb are drawn from externl ponnt p to a circle with center o prove that aobp are cyclic quardilateral

Answer 1

Note : The figure above is for 2) problem .

1)First question: Find the value of m so that the quadratic equation mx(x-7)49=0 has two equal roots .

ANSWER :

$given \: mx(x - 7)49 = 0 \\ mx(x - 7)49 = 0 \\49 mx {}^{2} - 343mx = 0 \\ since \: the \: equation \: has \: equal \: roots \: so \: \\ \\ {(343m})^{2} - 4 \times 49m \times 0 = 0 \\ ( {343m})^{2} = 0 \\ m = 0$
2)Again , PA and PB are drawn from external to a circle with centre O . Prove that AOPB is a cyclic quadrilateral.

See the figure :

We know that the radius and the tangent at point of contact are perpendicular to each other ,so

<OAP=<OBP=90°

Now from Quadrilateral AOBP

<OAP+<OBP +<AOB+<APB=360°
90°+90°+<AOP+<APB=360°
<AOP+<APB=180°

Thus AOBP is cyclic quadrilateral

HOPE IT WOULD HELP