If (h,k) is the centre of the circle touches y−axis at a distance of 12units from the origin and makes an intercept of 10units on x−axis, then the equation of circle for which (h+k) is minimum, is
A
(x−13)2+(y+12)2=169
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(x+13)2+(y−12)2=169
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(x−13)2+(y−12)2=169
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(x+13)2+(y+12)2=169
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D(x+13)2+(y+12)2=169 The centre of the circle is C≡(h,±12)
Assuming the point of intersection of the circle with x− axis at A and B, making a rough diagram, we get
In △ABC AC2=AM2+CM2⇒r2=122+52=169⇒r=13
Therefore the possible center of the circle are (±13,±12)
So, h+k is minimum when h=−13,k=−12
∴ The required equation of the circle is (x+13)2+(y+12)2=169