The circles $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ and $x^{2} + y^{2} + 4 x + 4 y - 1 = 0$

touch internally

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

touch externally

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

have

3 x + 4 y - 1 = 0

as the common tangent at the point of contact.

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

have

3 x + 4 y + 1 = 0

as the common tangent at the point of contact.

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

Open in App

Solution

The correct options are
B touch externally
C have $3 x + 4 y - 1 = 0$ as the common tangent at the point of contact.
For the given circles, $S_{1} = x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ and $S_{2} = x^{2} + y^{2} + 4 x + 4 y - 1 = 0$ ,

$C_{1} \equiv (1, 2), r_{1} = 2,$
and $C_{2} \equiv (- 2, - 2), r_{2} = 3$
Now, $r_{1} + r_{2} = 5$ and $C_{1} C_{2} = 5$ .
Hence, the circles touch externally. Also, the common tangent at the point of contact is $S_{1} - S_{2} = 0$ or $3 x + 4 y - 1 = 0$

Suggest Corrections

Similar questions

Q. The circles

x^{2} + y^{2} - 2 x - 4 y + 1 = 0

and

x^{2} + y^{2} + 4 x + 4 y - 1 = 0

Q. The circles

x^{2} + y^{2} + 4 x - 2 y + 4 = 0

and

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

Q. The circles

x^{2} + y^{2} - 2 x - 4 y - 20 = 0, x^{2} + y^{2} + 4 x - 2 y + 4

are

Find the equation of circle which touches $2 x - y + 3 = 0$ and pass through the points of intersection of the line $x + 2 y - 1 = 0$ and the circle $x^{2} + y^{2} - 2 x + 1 = 0$

Circles $x^{2} + y^{2} - 2 x - 4 y = 0 a n d x^{2} + y^{2} - 8 y - 4 = 0$

Join BYJU'S Learning Program

The circles x2+y2−2x−4y+1=0 and x2+y2+4x+4y−1=0

The circles $x^{2} + y^{2} - 2 x - 4 y + 1 = 0$ and $x^{2} + y^{2} + 4 x + 4 y - 1 = 0$