The standard equation of a circle is (x−h)^2+(y−k)^2=r^2, where the

Question

The standard equation of a circle is  (x−h)^2+(y−k)^2=r^2 , where the point (h,k) is the center of the circle and r represents the length of the radius. If a particular circle can be represented by the equation  x^2+y^2−8x+2y=−1 , what is the radius of that circle?

A. -1
B. 1
C. 2
D. 4
E. 8

Kurtosis · Accepted Answer

x^2 + y^2 - 8x + 2y = -1
To bring the above equation into its standard form, complete the squares by adding 16 and 1 on both the sides
x^2 - 8x + 16 + y^2 + 2y + 1 = -1 + 16 + 1
(x - 4)^2 + (y + 1)^2 = 16 --> This is in the form of (x-h)^2 + (y-k)^2 = r^2
r = 4

Answer: D

betancurj · Answer

Hi Bunuel,

This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.

I've been struggling trying to find out an easy procedure for this question. Can you help?

Thanks again

Juan

ENGRTOMBA2018 · Answer

Let me try to answer.

You are given that the standard equation of a circle ---> (x−h)^2+(y−k)^2=r^2 where, r=radius of the circle.

Now, you are given another circle with the equation ---> x^2 + y^2 - 8x + 2y = -1 , when you compare this to the standard equation of the circle, you should realize that it would be straightforward to calculate the radius of the circle if you can express x^2 + y^2 - 8x + 2y = -1 in the form of (x−h)^2+(y−k)^2=r^2

In order to make complete squares similar to (x-h)^2 or (y-k)^2, you need to remember that (x+a)^2=x^2+a^2+2ax and (x-a)^2=x^2+a^2-2ax

With these relations in mind, lets go back to x^2 + y^2 - 8x + 2y = -1 with a focus on converting x^2 + y^2 - 8x + 2y to 2 complete squares (1 with x and 1 with y).

x^2 + y^2 - 8x + 2y ---> (x^2-8x+16-16)+(y^2+2y+1-1) = (x-4)^2 -16 + (y+1)^2-1 ---> x^2 + y^2 - 8x + 2y = -1 can thus be written as :

(x-4)^2 -16 + (y+1)^2-1 = -1 --->(x-4)^2 + (y+1)^2 = 16 = 4^2.

NOw you can compare this form to (x−h)^2+(y−k)^2=r^2 to get h = 4, k=-1 and r=radius = 4.

Thus, D is the correct answer.

For questions such as these you need to move backwards from what has been given to you and to modify the given values into a more usable form.

Hope this helps.

OptimusPrepJanielle · Answer

Hi Juan,

The trick here is to convert everything in the standard form of equation of a circle, which is:
 (x−h)^2+(y−k)^2=r^2

We are given the equation of a circle in some other form (expanded form of the standard equation)

x^2+y^2−8x+2y=−1 
On grouping terms,

x^2−8x+y^2+2y=−1 
From here on, we need to make two perfect squares in x and y

The best way to understand which perfect square to make is by dissecting the coefficient of x and y
8 = 2*1*4, 2 = 2*1*1

We are writing the coefficients in this form, because the coefficient of x in  {(x+a)}^2 = x^2 + 2ax + a^2  (Eq i) is 2*1*a
 x^2−2*x*4 + y^2+2*1*y = -1

Now we need the constant terms to make these perfect squares.
The constant term in Eq (i) is a

Hence we can write  x^2−2*1*4x + y^2+2*1*y = -1  as
 x^2−2*x*4 + 16 + y^2+2*1*y + 1 = -1 + 16 + 1  (adding constants on both sides of the equation)
 {(x-4)}^2 + {(y+1)}^2 = 16 
 {(x-4)}^2 + {(y+1)}^2 = 4^2

Hence the radius = 4. Option D

betancurj · Answer

Thank you both, great help.

Nunuboy1994 · Answer

Bunuel  -

I simply factored (x-h)^2 + (y-k)^2 = r^2
 
   +   = r^2

If it can be reduced to

x^2 +y^2 -8x +2y -1

Then don't H and K have to be 4 and -1? Unless I'm making some assumption here that wouldn't necessarily apply to some other example- please let me know 
 
 If (x-4)^2 + (y - (-1))^2 = r^2
 (4-4)^2 + (1 +1)^2 = r^2
 (2)^2= r^2
 4= r^2
 4 = \sqrt{r} 
 4= r

Nunuboy1994 · Answer

Yes but don't forget to explain you derived 16 and 1 by taking half of the coefficients 8 and 1 and then squaring them. For example, if we have an algebraic expression that cannot be factored such as x-10x-18=0 - then we complete the square. How?

x-10x __ = 18 (add 18 to both sides or really just put 18 on the other side)
 x-10x +25 = 18 + 25 (remember how to get the coefficients? In this equation a= 1 b= 10; take (b/2)^2 and add to both sides
 (x-5)(x-5) = 43
 x-5 = + or minus 43

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

What should first jump out is that the equation for this particular circle does not match the standard equation of a circle. So you'll need to do some algebra to get to that point.

Your first step should be to factor the x and y terms so that they fit the standard form. Since each term needs to be factored into a squared parenthetical, you should use the common quadratic form  (a−b)^2=a^2−2ab+b^2  as your guide. Look at the coefficients −8x and +2y, recognizing that they need to fill the −2ab role in that common quadratic equation. This means that you'll factor the x and y terms into:

(x−4)^2=x^2−8x+16 
and
 (y−(−1)^2)=y^2+2y+1

So now you know that the left-hand side of the standard equation of a circle produces the combination of  x^2−8x+16+y^2+2y+1 , which equals  x^2−8x+y^2+2y+17 . So now you know two equations to be true:

x^2−8x+y^2+2y+17=r^2

and

x^2+y^2−8x+2y=−1

That second equation can be quickly made to look like the first by adding 1 to both sides:

x^2+y^2−8x+2y+1=0

Which means that the difference between the two equations is that the left-hand side is 16 apart and the right-hand side is r^2 apart. So to balance, add 16 to the left and r^2 to the right, meaning that r^2=16. This means that the radius is 4, making answer choice D correct.

vipulshahi · Answer

x^2 + y^2 -8x +2y = -1 
x^2 -8x +16 + y^2 + 2y +1 = -1+16+1
(x-4)^2 + (y+1)^2 = 4^2

We can compare the equation with (x-h)^2 + (y-k)^2 = r^2
we get, r=4

bumpbot · Answer

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