The equation of the circle touching the y - axis whose radius is 5 units and y-coordinate of the centre is -7- is A- x2-y2-10 x-14 y-49-0B- x2-y2-10 x-14 y-49-0C- x2-y2-10 x-14 y-25-0D- x 2- y 2-10 x -14 y -25-0

The correct options are A x^2+y^2 10x+14y+49=0 nbsp; B x^2+y^2+10x+14y+49=0Given : Circle touches the y axis and the radius is 5 units, so the centre of the c ...

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Standard XII

Mathematics

Definition of Circle

The equation ...

Question

The equation of the circle touching the $y -$ axis whose radius is $5$ units and $y -$ coordinate of the centre is $- 7,$ is

x^{2} + y^{2} - 10 x + 14 y + 49 = 0

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x^{2} + y^{2} + 10 x + 14 y + 49 = 0

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x^{2} + y^{2} + 10 x + 14 y + 25 = 0

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x^{2} + y^{2} - 10 x + 14 y + 25 = 0

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Solution

The correct options are
A $x^{2} + y^{2} - 10 x + 14 y + 49 = 0$
B $x^{2} + y^{2} + 10 x + 14 y + 49 = 0$
Given : Circle touches the $y -$ axis and the radius is $5 units$ , so the centre of the circle is
$C = (\pm 5, - 7)$
So, the equation of the circle is
$(x \pm 5)^{2} + (y + 7)^{2} = 5^{2}$
$\Rightarrow x^{2} + y^{2} - 10 x + 14 y + 49 = 0$
And $x^{2} + y^{2} + 10 x + 14 y + 49 = 0$

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The equation of the circle touching the y−axis whose radius is 5 units and y−coordinate of the centre is −7, is

The correct options are A x2+y2−10x+14y+49=0 B x2+y2+10x+14y+49=0Given : Circle touches the y−axis and the radius is 5 units, so the centre of the circle is C=(±5,−7) So, the equation of the circle is (x±5)2+(y+7)2=52 ⇒x2+y2−10x+14y+49=0 And x2+y2+10x+14y+49=0

The equation of the circle touching the $y -$ axis whose radius is $5$ units and $y -$ coordinate of the centre is $- 7,$ is