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

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

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

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

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

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

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

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Solution

The correct option is D $y^{2} - 10 x - 6 y + 14 = 0$
Given the equation of circle is $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$

We know $c = f^{2}$

center $(3, 3)$ .....From equation

and $(g - 3)^{2} + (f - 3)^{2} = [\sqrt{9 + 9 - 14} + \sqrt{g^{2} + f^{2} - c^{2}}]^{2}$

$\Rightarrow g^{2} + f^{2} - 6 g - 6 f + 18 = (2 + g)^{2}; as c = f^{2}$

$\Rightarrow f^{2} - 10 g - 6 f + 14 = 0$

Therefore, locus of centre $(g, f)$ is $y^{2} - 10 x - 6 y + 14 = 0$ .

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Similar questions

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

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 the centre of the 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 equations

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:

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

The correct option is D y2−10x−6y+14=0Given the equation of circle is x2+y2−6x−6y+14=0We know c=f2center(3,3).....From equationand (g−3)2+(f−3)2=[√9+9−14+√g2+f2−c2]2⇒g2+f2−6g−6f+18=(2+g)2;as c=f2⇒f2−10g−6f+14=0Therefore, locus of centre (g,f) is y2−10x−6y+14=0.

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