The equation of the circle with centre at 0-0 and touching the line 3 x -4 y -5-0 is A- x2-y2-1B- x2-y2-4C- x2-y2-9D- x 2- y 2-16

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

Position of a Line with Respect to Circle

The equation ...

Question

The equation of the circle with centre at $(0, 0)$ and touching the line $3 x + 4 y + 5 = 0$ is

x^{2} + y^{2} = 1

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

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

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

x^{2} + y^{2} = 9

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

x^{2} + y^{2} = 16

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

Open in App

Solution

The correct option is A $x^{2} + y^{2} = 1$
The radius of the circle is the perpendicular distance from the centre $(0, 0)$ to the line $3 x + 4 y + 5 = 0$
$r = \frac{3 \times 0 + 4 \times 0 + 5}{\sqrt{3^{2} + 4^{2}}} = 1$
Hence the equation of the circle is given by
$x^{2} + y^{2} = 1$

Suggest Corrections

Similar questions

Q. Equation of the circle, centred at

(1, - 5)

and touching the line

3 x + 4 y = 8,

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,

Find the equation of circle circumscribing the quadrilateral formed by four lines $3 x + 4 y - 5 = 0$ , $4 x - 3 y - 5 = 0$ , $3 x + 4 y + 5 = 0$ and $4 x - 3 y + 5 = 0$

The radius of the circle S is same as the radius of $x^{2} + y^{2} - 2 x + 4 y - 11 = 0$ and the centre of S is the centre of $x^{2} + y^{2} - 2 x - 4 y + 11 = 0$ . Find the equation of S.

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

Join BYJU'S Learning Program

The equation of the circle with centre at (0,0) and touching the line 3x+4y+5=0 is

The correct option is A x2+y2=1The radius of the circle is the perpendicular distance from the centre (0,0) to the line 3x+4y+5=0 r=3×0+4×0+5√32+42=1 Hence the equation of the circle is given by x2+y2=1

The equation of the circle with centre at $(0, 0)$ and touching the line $3 x + 4 y + 5 = 0$ is

The correct option is A $x^{2} + y^{2} = 1$
The radius of the circle is the perpendicular distance from the centre $(0, 0)$ to the line $3 x + 4 y + 5 = 0$
$r = \frac{3 \times 0 + 4 \times 0 + 5}{\sqrt{3^{2} + 4^{2}}} = 1$
Hence the equation of the circle is given by
$x^{2} + y^{2} = 1$