# Angle between tangentsdrawn from a point p to circle x2+y2-4x-8y+8=0 is 600 then length of chord of contact of p is

Samyak Jain
333 Points
4 years ago
This is a simple question if you know some properties of circle.
I can’t explain it diagrammatically here. Please understand. I’ve explained it fully.

Let the centre of circle be O and tangents from P meet the circle at Q and R. Let M be the point of
intersection of QR and OP. Then QR is the chord of contact of P.

Radius of circle, r = $\dpi{80} \sqrt{g^2+f^2-c}$ = $\dpi{80} \sqrt{(-2)^2+(-4)^2-8}$ = $\dpi{80} \sqrt{12}$ = 2$\dpi{80} \sqrt{3}$
We know at the point of contact, radius of circle is perpendicular to tangent.
So, $\dpi{100} \angle$OQP = $\dpi{100} \angle$ORP = 90$\dpi{80} \degree$.
Also, QR is perpendicular to OP i.e. $\dpi{100} \angle$OMQ = 90$\dpi{80} \degree$.

OP bisects $\dpi{100} \angle$QPR  i.e.  $\dpi{100} \angle$QPO = $\dpi{100} \angle$RPO = 30$\dpi{80} \degree$.
$\dpi{100} \therefore$ $\dpi{100} \angle$QOP = 60$\dpi{80} \degree$.
QM = OQsin60$\dpi{80} \degree$ = rsin60$\dpi{80} \degree$ = 2$\dpi{80} \sqrt{3}$.($\dpi{80} \sqrt{3}$ / 2) = 3.
OP bisects QR i.e. QM = MR  or
QR = 2.QM = 2.3  =  6 unit = length of chord of contact of P.