Q.

The equation of the smallest circle passing through the points of intersection of the line x+y−1=0  and the circle x2+y2=9

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

x2+y2+x+y−8=0

b

x2+y2−x−y−8=0

c

x2+y2−x+y−8=0

d

x2+y2−x−y+8=0

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given circle x2+y2=9.......(1)  and line if x+y−1=0............(2) The point of intersection of (1)&(2)   is (1)+λ(2)=0 ⇒x2+y2−9+λ(x+y−1)=0 ⇒x2+y2+λx+λy−λ−9=0.........(3) Centre (−λ2,−λ2) Since equation (3) is a smallest circle then equation (2)  is a diameter of the circle⇒Centre(−λ2,−λ2) lies on (2) −λ2,−λ2−1=0⇒λ=−1 Substituting λ=−1  in equation (3) ⇒x2+y2−9−(x+y−1)=0 ⇒x2+y2−x−y−8=0
Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
The equation of the smallest circle passing through the points of intersection of the line x+y−1=0  and the circle x2+y2=9