If (-2,5) is the centre of the circle which touches y-axis, then the equation of the circle is

x²+y²-4x-10y+25=0

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

x²+y²-4x-10y+4=0

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

x²+y²+4x+10y+2=0

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

x²+y²+4x-10y+25=0

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

Open in App

Solution

The correct option is D
x²+y²+4x-10y+25=0

C =(-2,5)
Circle touches Y-axis \
$∴$ r = 1 - 3 = 2
$(x + 2)^{2} + (y - 5)^{2}$ = $2^{2}$
$x^{2} + y^{2} + 4 x - 10 y + 25$ = 0

Suggest Corrections

Similar questions

(- 2, 5)

y

(2, 0)

(3, 4)

y

2 x - y + 1 = 0

(2, 5)

x + y - 9 = 0

x^{2} + y^{2} - 6 x - 6 y + 14 = 0

Join BYJU'S Learning Program