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I tried squaring both sides to get this $y^2 = 24-2x-x^2$, then putting the $x$'s with the $y$'s to get $y^2 + x^2 + 2x = 24$. Then I tried dividing everything by 24, but I don't see it. Tried factoring too.

kba
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J L
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3 Answers3

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$$y=\sqrt{24-2x-x^2}\\y^2=24-2x-x^2\\y^2=25-(x+1)^2\\(x+1)^2+y^2=5^2$$

From the second to the third line, I completed the square. Notice now that we have the equation of a circle centered at $(-1,0)$ of radius $5$. However, your original equation only represents the top half of this circle, because $y$ is always positive.

Here is a picture of the graph:

enter image description here

Jared
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equation of circle in general form is

$$ x^2 + y^2 + (2\times g \times x) + (2\times f \times y) + c = 0 $$

$$\boxed{Radius = \sqrt{g^2 + f^2 -c }}$$

compare this with your equation

$$ x^2 + y^2 + (2\times 1 \times x) + (2\times 0 \times y) + (-24) = 0 $$ here g = 1 , f =0 , c = -24

so radius = $ \sqrt { 1^2 +0^2 -(-24)} $

= $\sqrt {25 } $ = 5

From above way the tedious job of factorizing is eliminated . The equation given in your case is equation in general form Why I suggested this solution is that sometimes it is tedious to change a equation from general form $ x^2 + y^2 + (2\times g \times x) + (2\times f \times y) + c = 0 $ , to standard form $ (x-a)^2 + (y-b)^2 = (radius)^2 $, in those cases the radius formula given above will be helpful .

Harish Kayarohanam
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the equation was; x^2 + y^2 + 2x -24=0 is in the general form, and the general equation of the circle is ax^2 + ay^2 + dx + ey + f=0; using the equation alone we can solve for the radius,center,area and circum.|| Using the formulas.:|| center: h= -d/2a , k=-e/2a|| radius; r^2= (d^2+ e^2 -4af)/4a^2|| area ; A= pi(r^2)||

applying:|| given: a=1 d=2 e=0 f=-24|| center; h=(-2)/2(1)=-1 k=-(0)/2(1)=0|| radius; r^2=(2^2 + 0^2 - 4(1)(-24))/4(1^2)|| r^2=25 = 5 ||

silentjoker
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