1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines x2−y2−2x+4y−3=0, is

A

x2+y22x4y+4=0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

x2+y2+2x+4y4=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

x2+y22x+4y+4=0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
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 A x2+y2−2x−4y+4=0 Let the required equation of the circle be (x−h)2+(y−k)2=a2 Comparing the given equation x2−2x+4y−3=0 with ax2+by2+2hxy+2gx+2fy+c=0 we get : a=1,b−1,h=0,g=−1,f=2,c=−3 Intersection point =(hf−bgab−h2,gh−afab−h2)=(−1−1,−2−1)=(1,2) Thus, the centre of the circle is (1, 2). The equation of the required circle is (x−1)2+(y−2)2=a2 Since circle passes through (1, 1), we have 1=a2 ∴ Equation of the required circle : (x−1)2+(y−2)2=1 ⇒x2+y2−2x−4y+4=0

Suggest Corrections
3
Join BYJU'S Learning Program
Related Videos
Definition and Standard Forms
MATHEMATICS
Watch in App
Join BYJU'S Learning Program