Find a relationship between $x$ and $y$ so that the distance between the points $(x, y)$ and $(- 2, 4)$ is equal to $5$ .

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

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x^{2} + y^{2} + 4 x - 8 y - 5 = 0

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x^{2} + y^{2} + 4 x - 8 y - 6 = 0

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x^{2} + y^{2} + 4 x - 8 y - 53 = 0

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Solution

The correct option is B $x^{2} + y^{2} + 4 x - 8 y - 5 = 0$
Using distance formula :

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

$∴ \sqrt{(x + 2)^{2} + (y - 4)^{2}} = 5$ ...{given}

By squaring both the sides we get,

$(x + 2)^{2} + (y - 4)^{2} = 25$ ,

Now by expanding the squares we get,

$x^{2} + 4 + 4 x + y^{2} + 16 - 8 y = 25$

$\Rightarrow x^{2} + y^{2} + 4 x - 8 y - 5 = 0$

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Similar questions

The equation of the circle which intersects circles $x^{2} + y^{2} + x + 2 y + 3 = 0, x^{2} + y^{2} + 2 x + 4 y + 5 = 0$ and

$x^{2} + y^{2} - 7 x - 8 y - 9 = 0$ at right angle, will be

2 y = 4 x - 6

y^{2} = - 16 x

A

B

A B

S_{1} : x^{2} + y^{2} - 4 x + 6 y + 11 = 0

S_{2} : x^{2} + y^{2} - 2 x + 8 y + 13 = 0

y^{2} - 4 x - 8 y + 40 = 0

x^{2} - 8 x - 4 y + 40 = 0

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