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A circle has to points on the circumference (0, 1) and (0, 9) that bisect the circle and I have to give the equation of the circle in general form.

So I can find the midpoint of the circle as ${(0, {{9 + 1} \over 2})}$ = (0, 5).

Using the distance formula to find the distance between the two points I get ${\sqrt {(0)^2 + (9-1)^2}}$ which is 8, divided by 2 is 4.

So I would give the equation as:

${x^2 + y^2 -10y + 16 = 0}$

But the book says

${x^2 + y^2 -10y +9 = 0}$

I don't get how the radius is 3.

dagda1
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    I think you need to complete the square on your $y$ term with a constant of $25$. Then the difference between that and the radius squared will give you the 9 left over. – turkeyhundt Sep 23 '15 at 19:53

4 Answers4

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Your book is right. Consider a circumference of radius 4 centered at $(0,0)$. Its equation is:

$$ x^2 + y^2 = 4^2 = 16 $$

Now, with your circumference centered at $(0,5)$ we get:

$$ x^2 + (y-5)^2 = 16 $$

$$ x^2 + y^2 - 10y + 25 - 16 = 0 $$

$$ x^2 + y^2 - 10y + 9 = 0 $$

Dleep
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From the equation for an ellipse:

$(\frac{x-x_c}{a})^2 + (\frac{y-y_c}{b})^2 = 1 $

(Here, a = b = r since it is a circle)

You found the radius to be 4, which is correct. However, if you examine where the center of the circle is -- $(0,5)$, you get the following equation:

$(x-0)^2+(y-5)^2 = 4^2$

$x^2+y^2-10y+25 = 16$

or

$x^2+y^2-10y+9 = 0$

costrom
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Equation of Circle whose center is at $C(h,k)$ and radius $r$ is $(x-h)^2+(y-k)^2=r^2$

So here $C(h,k) = C(0,5)$ and $r=4$

So equation is $$(x-0)^2+(y-5)^2=4^2\Rightarrow x^2+y^2+25-10y = 16$$

So we get $$x^2+y^2-10y+9=0$$

juantheron
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Notice, You have correctly found the center of circle $(0, 5)$ & the radius $4$. Probably you have done the mistake while finding out the equation of circle.

The equation of the circle with the center $(0, 5)$ & the radius $4$ is written as follows $$(x-0)^2+(y-5)^2=(4)^2$$ $$x^2+y^2-10y+25=16$$ $$\color{red}{x^2+y^2-10y+9=0}$$ so the answer in the book is correct.

Harish Chandra Rajpoot
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