The equation of the circle of radius 3 that lies in the second quadrant and touching the lines x = 0 and y = 0 is

- 6x + 6y + 9 = 0

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

- 6x - 6y + 9 = 0

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

+ 6x - 6y + 9 = 0

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

+ 6x + 6y + 9 = 0

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

Open in App

Solution

The correct option is D + + 6x + 6y + 9 = 0
The required circle equation is $(x + 3)^{2}$ + $(y + 3)^{2}$ = $3^{2}$
$\Rightarrow$ $x^{2}$ + $y^{2}$ + 6x + 6y + 9 = 0

Suggest Corrections

Similar questions

Q. Equation of a circle which passes through (3, 6) and touches the axes is
(a) x² + y² + 6x + 6y + 3 = 0
(b) x² + y² – 6x – 6y – 9 = 0
(c) x² + y² – 6x – 6y + 9 = 0
(d) None of these

Q. The equation of the image of the circle

x^{2} + y^{2} - 6 x - 4 y = 0

in the bisector of

2^{nd}

and

4^{th}

quadrant is

Q. From the point A (0, 3) on the circle

x^{2} + 4 x + (y - 3)^{2} = 0

a chord AB is drawn and extended to a point M such that AM =2AB. The equation of the locus of M is___

Q. The locus of the centre of a circle, which touches externally the circle

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

and also touches the y-axis, is given by the equation

Tangents are drawn from the point P(1, 8) to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B, then equation of the circumcircle of the triangle PAB is

Join BYJU'S Learning Program

The equation of the circle of radius 3 that lies in the second quadrant and touching the lines x = 0 and y = 0 is

The correct option is D + + 6x + 6y + 9 = 0The required circle equation is (x+3)2 + (y+3)2 = 32 ⇒ x2 + y2 + 6x + 6y + 9 = 0

The correct option is D + + 6x + 6y + 9 = 0
The required circle equation is $(x + 3)^{2}$ + $(y + 3)^{2}$ = $3^{2}$
$\Rightarrow$ $x^{2}$ + $y^{2}$ + 6x + 6y + 9 = 0