SATHEE: Circle 4 Question 9

Circle 4 Question 9

9. 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 $(1993, 1 M)$

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

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

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

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Answer:

Correct Answer: 9. (d)

Solution:

Let $(h, k)$ be the centre of the circle which touches the circle $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ and $Y$ -axis.

The centre of given circle is $(3, 3)$ and radius is $\sqrt{3^{2} + 3^{2} - 14} = \sqrt{9 + 9 - 14} = 2$

Since, the circle touches $Y$ -axis, the distance from its centre to $Y$ -axis must be equal to its radius, therefore its radius is $h$ . Again, the circles meet externally, therefore the distance between two centres $=$ sum of the radii of the two circles.

Hence,

$(h - 3)^{2} + (k - 3)^{2} = (2 + h)^{2}$

$h^{2} + 9 - 6 h + k^{2} + 9 - 6 k = 4 + h^{2} + 4 h$

i.e.

$k^{2} - 10 h - 6 k + 14 = 0$

Thus, the locus of $(h, k)$ is

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

Circle 4 Question 9

9. 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 $(1993, 1 M)$

Answer:

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Circle 4 Question 9

9. 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 (1993,1M)

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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9. 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 $(1993, 1 M)$