Equation of Family of Circles Passing through Points of Intersection of Circle and a Line
The equation ...
Question
The equation of circle passing through the origin and point of intersection of circle x2+y2−2x+4y−20=0 and line x+y−1=0 is
A
x2+y2+22x−16y=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x2+y2+22x+16y=0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x2+y2−22x−16y=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Ax2+y2−22x−16y=0 As the equation of circle passes through point of intersection of circles x2+y2−2x+4y−20=0 and line x+y−1=0
Let the equation of required circle is (x2+y2−2x+4y−20)+λ(x+y−1)=0 Since, it passes through origin. Therefore, −20−λ=0⇒λ=−20 Hence, the required equation of circle is x2+y2−22x−16y=0.