The equation of the circle which touches x-axis and whose centre is 1-2- is - x2-y2-2 x-4 y-1-0- x2-y2-2 x-4 y-1-0- x2-y2-2 x-4 y-1-0D- x2-y2-4 x-2 y-4-0

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

Distance Formula

The equation ...

Question

The equation of the circle which touches x-axis and whose centre is (1, 2), is

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

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

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

Open in App

Solution

The correct option is B
Centre (1, 2) and since circle touches x-axis, therefore, radius is equal to 2.
Hence the equation is $(x - 1)^{2} + (y - 2)^{2} = 2^{2}$
$\Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
Trick: The only circle is $x^{2} + y^{2} - 2 x - 4 y + 1 = 0,$ whose centre is (1, 2).

Suggest Corrections

Similar questions

Q. The equation of the circle with centre at (1, -2) and passing through the centre of the given circle

x^{2} + y^{2} + 2 y - 3 = 0,

Q. The locus of midpoint of chord of the circle

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

, which makes an angle of

120^{\circ}

at the centre, is

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

Q. Equation of the circle, centred at

(1, - 5)

and touching the line

3 x + 4 y = 8,

The equation of a circle which cuts the three circles

$x^{2} + y^{2} - 3 x - 6 y + 14 = 0,$

$x^{2} + y^{2} - x - 4 y + 8 = 0$

$x^{2} + y^{2} + 2 x - 6 y + 9 = 0$

orthogonally is $- ------------- -$

Join BYJU'S Learning Program

The equation of the circle which touches x-axis and whose centre is (1, 2), is

The correct option is B Centre (1, 2) and since circle touches x-axis, therefore, radius is equal to 2. Hence the equation is (x−1)2+(y−2)2=22 ⇒x2+y2−2x−4y+1=0 Trick: The only circle is x2+y2−2x−4y+1=0, whose centre is (1, 2).

The correct option is B
Centre (1, 2) and since circle touches x-axis, therefore, radius is equal to 2.
Hence the equation is $(x - 1)^{2} + (y - 2)^{2} = 2^{2}$
$\Rightarrow x^{2} + y^{2} - 2 x - 4 y + 1 = 0$
Trick: The only circle is $x^{2} + y^{2} - 2 x - 4 y + 1 = 0,$ whose centre is (1, 2).