# What is the centre and radius of a circle with equation x² + y² - 8x - 4y = 9?

• Crosshash
I didn't know whether it was completely correct.In summary, the equation of a circle is x² + y² - 8x - 4y = 9. By expanding the brackets, it can be simplified to (x-4)² + (y-2)² = 9. The correct expansion is x² - 8x + y² - 4y + 20 = 9. By equating coefficients and solving for a and b, the centre of the circle is determined to be (4,2). The radius of the circle is then found to be √29.
Crosshash

## Homework Statement

A circle has equation x² + y² - 8x - 4y = 9

(i) Show that the centre of this circle is (4,2) and find the radius of the circle.

## Homework Equations

Circle equation = (x-a)² + (x-b)² = r²

## The Attempt at a Solution

Well, if the centre of the circle is (4,2). Then the equation will be something like:

(x-4)² + (y-2)² = 9

except, if I expand out the two brackets, I get

(x² -8x + 16) + (y² -4y + 4) = 9

x² - 8x + y² -4y + 20 = 9

Here is where i get confused

So (i'm just double checking here since I don't have the answers), is the radius of the circle sqr9 or sqr 11? And is this really showing that (4,2) is the centre?

Thanks

Last edited:
Crosshash said:
Well, if the centre of the circle is (4,2). Then the equation will be something like:

(x-4)² + (y-2)² = 9

except, if I expand out the two brackets, I get

(x² -4x + 16) + (y² -2y + 4) = 9

x² - 4x + y² -2y + 20 = 9

Here is where i get confused
You have expanded the brackets incorrectly.

Hootenanny said:
You have expanded the brackets incorrectly.

oops, fix'd (i hope :) )

Crosshash said:

## Homework Statement

A circle has equation x² + y² - 8x - 4y = 9

(i) Show that the centre of this circle is (4,2) and find the radius of the circle.

## Homework Equations

Circle equation = (x-a)² + (x-b)² = r²

## The Attempt at a Solution

Well, if the centre of the circle is (4,2). Then the equation will be something like:

(x-4)² + (y-2)² = 9

except, if I expand out the two brackets, I get

(x² -8x + 16) + (y² -4y + 4) = 9

x² - 8x + y² -4y + 20 = 9

Here is where i get confused

So (i'm just double checking here since I don't have the answers), is the radius of the circle sqr9 or sqr 11? And is this really showing that (4,2) is the centre?

Thanks

Don't let the post above you put you off. You've expanded right but your fundamentally wrong.

I'll explain this simply.

x² + y² - 8x - 4y = 9

ok you have to make the factorization to reach the -8x and the -4y

which is (x-4)^2 + (y-2)^2 = 9 which is correct.

Though this isn't finished, because when you factorized into those brackets you also added an extra -4^2 and a -2^2 that you didn't need. So to balance this, you have to add these to the right hand side of the equation.

it ends up : (x-4)^2 + (y-2)^2 = 9 + 4^2 + 2^2
= 29

DeanBH said:
Don't let the post above you put you off. You've expanded right but your fundamentally wrong.
No he didn't expand the brackets correctly, notice that he edited his post after I posted. See my quoted text above.

Hootenanny said:
No he didn't, notice that he edited his post after I posted. See my quoted text above.

I thought you told him he expanded correctly.

Thanks for the replies, I managed to reach sqr29 using a different method though.

x² + y² - 8x - 4y = 9

so

x² + y² - 8x - 4y - 9 = 0

and the equation of a circle is

(x - a)² + (x - b)² = r²

expand out

x² + y² - 2ax - 2by + a² + b² - r²

equating coeficients

-2a = -8
a = 4

-2b = -4
b = 2

constant terms

a² + b² - r² = -9

4² + 2² - r² = -9

-r² = -29

r = sqr29

and I think that shows that the centre is (4,2) as well

This way seems logical but much longer

## 1. What is the standard form of a circle equation?

The standard form of a circle equation is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle and r represents the radius.

## 2. How do you find the center and radius of a circle given its equation?

To find the center and radius of a circle given its equation, first rewrite the equation in standard form. Then, the values of h and k in the equation represent the x-coordinate and y-coordinate of the center, respectively. The value of r can be found by taking the square root of the number on the other side of the equal sign.

## 3. Can a circle equation have a negative radius?

No, a circle equation cannot have a negative radius. The radius represents the distance from the center of the circle to any point on the circle's circumference, and distance cannot be negative.

## 4. How do you graph a circle given its equation?

To graph a circle given its equation, first find the center and radius using the method described in question 2. Then, plot the center point on the coordinate plane and use the radius to mark points on the circle's circumference in all four directions (up, down, left, right). Finally, connect the points to form the circle.

## 5. Can a circle equation have a non-integer radius?

Yes, a circle equation can have a non-integer radius. The radius can be any real number, as long as it is positive. This means that a circle can have a radius of 1.5, 2.7, or any other non-integer value.

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