From a point P(2,2√2) tangents PO and PR, are drawn to the circle x2+y2=a2. If OR is touching the circle x2+y2=3, then value of 'a'
A
4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2√3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3√2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
√6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D√6 eqn of chord of contact from p(2,2√2) to x2+y2=a2is T=0⇒2x+2√2y=a2---{1} LineR:2x+2√2y=a2 then, line oR is also a tongent for x2+y2=3 is then r=√3=|0+0−a2√4+8| 2×3=|a2| ⇒a=±√6