Question

${x}^{2} + {y}^{2} = 40$
Find x and y​

Answer 1

Hello Mate,

Here is your answer,

In this case we can take any value for 'y' and solve for 'x', or any values of 'x' and solve for 'y', hence this equation has infinite number of solutions.

To make this clear, let's take any value of 'x' or 'y'.

Let x = 0,

Then

${x}^{2} + {y}^{2} = 40 \\ \\ 0 + {y}^{2} = 40 \\ \\ y = \sqrt{40} \\ \\ y = 2 \sqrt{10}$

This is one of the solution. Now let's take any value for 'y',

Let y = 2,

${x}^{2} + {y}^{2} = 40 \\ \\ = {x}^{2} + 4 = 40 \\ \\ {x}^{2} = 36 \\ \\ x = 6$

This is another solution (6, 2)

So taking any values of 'x' and 'y' we can get infinite solutions.

Hope it helps:)

Answer 2

$x {}^{2} + y {}^{2} = 40$

....(1)

it is a equation of circle

whose centre is origin.

❤️ we know that

general equation of circle is

x^ 2 + y^2 = R^ 2 ..... (2)

whose centre is (0,0)

therefore ,

radius of circle is R= √40 cm = 6.324 cm

[ on comparing equation (1) and (2)]

I hope it helps you

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