The locus of the centre of a circle which touch the circles $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ externally and also the Y - axis is given by

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

Open in App

Solution

The correct option is D

Let (x, y) be centre of circle which touch y - axis and given circle.
At any point, the radius of this circle will be equal to 'x' units. Since this circle is touching the given circle which has a radius of two units,
$∴$ Distance between centres of the two circles = 2 + x

$\Rightarrow \sqrt{(x - 3)^{2} + (y - 3)^{2}} = 2 + x$

$y^{2} - 10 x - 6 y + 14 = 0$

Suggest Corrections

Similar questions

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

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

externally and touches the y-axis s given by:

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

Q. The locus of a 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

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

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:

Join BYJU'S Learning Program

The locus of the centre of a circle which touch the circles x2+y2−6x−6y+14=0 externally and also the Y - axis is given by

The locus of the centre of a circle which touch the circles $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ externally and also the Y - axis is given by