0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# Equation of the circle which touches the coordinate axes and passes through the point (2,1) is ?

A
x2+y2+8x+8y29=0
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B
x2+y210x10y+25=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x2+y2+10x+10y35=0
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
x2+y220x20y+55=0
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is B x2+y2−10x−10y+25=0Let the required equation be x2+y2+2gx+2fy+c=0It passes through the point (2,1)⟹4+1+4g+2f+c=04g+2f+c+5=0⋯(1)It touches x-axis⟹g2−c=0g2=c⋯(2)It touches y-axis⟹f2−c=0f2=c⋯(3)From (2),(3)g2=f2⟹g=f⋯(4)Put (2)(4) in (1)⟹4g+2g+g2+5=0g2+6g+5=0(g+1)(g+5)=0g=−1,−5Put g=−5⟹f=−5⟹c=25So the required Equation is x2+y2−10x−10y+25=0

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Family of Circles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program