Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
        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
28 days ago

Samyak Jain
325 Points
							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 ofintersection 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  orQR = 2.QM = 2.3  =  6 unit = length of chord of contact of P.

24 days ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 53 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

Post Question