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Equation of Circle with Origin as Center

Without chang...

Question

Without changing the direction of the axes, the origin is transferred to the point $(2, 3)$ . Then the equation $x^{2} + y^{2} - 4 x - 6 y + 9 = 0$ changes to

A

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

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B

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

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C

x^{2} + y^{2} = 4, x^{2} + y^{2} - 8 x - 12 y + 48 = 0

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D

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

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Solution

The correct option is B $x^{2} + y^{2} = 4$
$x \to x + 2, y \to y + 3$
Given equation of circle is $x^{2} + y^{2} - 4 x - 6 y + 9 = 0$
$∴ {(x + 2)}^{2} + {(y + 3)}^{2} - 4 (x + 2) - 6 (y + 3) + 9 = 0$
$\Rightarrow x^{2} + 4 x + 4 + y^{2} + 6 y + 9 - 4 x - 8 - 6 y - 18 + 9 = 0$
$\Rightarrow x^{2} + y^{2} - 4 = 0$

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Similar questions

Q. Without changing the direction of the axes, the origin is transferred to the point

(2, 3)

. Then the equation

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

Circles	Number of common tangents
I: $x^{2} + y^{2} = 4$ $x^{2} + y^{2} - 8 x + 12 = 0$	a) $0$
II: $x^{2} + y^{2} = 1$ $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$	b) $1$
III: $x^{2} + y^{2} = 16$ $x^{2} + y^{2} - 8 x + 6 y - 56 = 0$	c) $4$
IV: $x^{2} + y^{2} - 2 x - 6 y + 9 = 0$ $x^{2} + y^{2} + 6 x - 2 y + 1 = 0$	d) $3$

A B

is drawn from the point

A (0, 3)

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

and is extended to

M

A M = 2 A B

M

Q. I : The equations to the direct common tangents to the circles

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

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

y - 1 = 0

4 x - 3 y - 9 = 0

II : The equations to the transverse common tangents to the circles

x^{2} + y^{2} - 4 x - 10 y + 28 = 0

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

x - 1 = 0, 3 x + 4 y - 21 = 0

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___

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