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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8033
GMAT 1: 760 Q51 V42 GPA: 3.82
(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 67% (01:56) correct 33% (02:38) wrong based on 66 sessions

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(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and center (a,b). What is the radius of the circle with equation x^2 + y^2 – 10x + 4y = 20?

A. 4
B. 5
C. 6
D. 7
E. 8

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Intern  B
Joined: 07 Feb 2018
Posts: 12
Location: India
Concentration: Technology, General Management
Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and  [#permalink]

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2
1
The secret to solving such questions is to make the given equation in question similar to the definition of circle.

The second key is to see the terms with single power (10x and 4y).

Now lets solve......

Make the equation as...

x^2 -(2 * x *5) + 5^2 - 5^2 + y^2 + (2 * y *2 ) + 2^2 -2^2 = 20
which means
(x-5)^2 + (y+2)^2 = 20+ 5^2+2^2 = 49 = 7^2

Thus 7 is the radius....

Give Kudos if you like the solution...Cheers !!! Happy Learning !!!
Manhattan Prep Instructor G
Joined: 04 Dec 2015
Posts: 832
GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and  [#permalink]

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MathRevolution wrote:
(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and center (a,b). What is the radius of the circle with equation x^2 + y^2 – 10x + 4y = 20?

A. 4
B. 5
C. 6
D. 7
E. 8

This is an interesting problem, because while I've seen 'completing the square' used in solutions for GMAT problems before, I'm not sure that I've seen it required in an official problem. I'd be curious to see if anybody can show an official problem that requires you to do this.

Here's how it works: you need to take the equation you have, and put it in a form so that all of the variables are contained in the $$(x-a)^2$$ and the $$(y-b)^2$$ parts of the equation, and you only have numbers left over.

Focus on the x's first. The terms that have an x in them are $$x^2 - 10x$$. You want to make that look like (x-something)^2.

Well, when you square (x-a), you end up with $$x-2ax+a^2$$. So, 2a has to equal 10 in order for the expressions to look alike. That means we're looking at $$(x-5)^2$$.

$$(x-5)^2$$ equals$$x^2 - 10x + 25$$. Therefore, $$x^2 - 10x$$ can be rewritten as $$(x-5)^2-25$$.

Similarly, for the y terms, we'll rewrite as $$(y+2)^2$$, which equals $$y^2 + 4y + 4$$. Therefore, $$y^2+4y$$ can be rewritten as$$(y+2)^2-4$$.

Now, put those back into the original equation.

$$x^2 + y^2 – 10x + 4y = 20$$
$$(x-5)^2 - 25 + (y+2)^2 - 4 = 20$$
$$(x-5)^2 + (y+2)^2 = 20 + 4 + 25 = 49 = 7^2$$

So, the radius is 7.
_________________ Chelsey Cooley | Manhattan Prep | Seattle and Online

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8033
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and  [#permalink]

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=>

$$x^2 + y^2 – 10x + 4y = 20$$
$$=> x^2 – 10x + y^2 + 4y = 20$$
$$=> x^2 – 10x + 25 + y^2 + 4y + 4 = 20 + 25 + 4$$
$$=> (x-5)^2 + (y+2)^2 = 49$$
$$=> (x-5)^2 + (y+2)^2 = 7^2$$

Thus the circle has center $$(5,-2)$$ and radius $$7$$.

Therefore, the answer is D.
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Senior Manager  S
Joined: 12 Sep 2017
Posts: 302
Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and  [#permalink]

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ccooley wrote:
MathRevolution wrote:
(x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and center (a,b). What is the radius of the circle with equation x^2 + y^2 – 10x + 4y = 20?

A. 4
B. 5
C. 6
D. 7
E. 8

This is an interesting problem, because while I've seen 'completing the square' used in solutions for GMAT problems before, I'm not sure that I've seen it required in an official problem. I'd be curious to see if anybody can show an official problem that requires you to do this.

Here's how it works: you need to take the equation you have, and put it in a form so that all of the variables are contained in the $$(x-a)^2$$ and the $$(y-b)^2$$ parts of the equation, and you only have numbers left over.

Focus on the x's first. The terms that have an x in them are $$x^2 - 10x$$. You want to make that look like (x-something)^2.

Well, when you square (x-a), you end up with $$x-2ax+a^2$$. So, 2a has to equal 10 in order for the expressions to look alike. That means we're looking at $$(x-5)^2$$.

$$(x-5)^2$$ equals$$x^2 - 10x + 25$$. Therefore, $$x^2 - 10x$$ can be rewritten as $$(x-5)^2-25$$.

Similarly, for the y terms, we'll rewrite as $$(y+2)^2$$, which equals $$y^2 + 4y + 4$$. Therefore, $$y^2+4y$$ can be rewritten as$$(y+2)^2-4$$.

Now, put those back into the original equation.

$$x^2 + y^2 – 10x + 4y = 20$$
$$(x-5)^2 - 25 + (y+2)^2 - 4 = 20$$
$$(x-5)^2 + (y+2)^2 = 20 + 4 + 25 = 49 = 7^2$$

So, the radius is 7.

Hello ccooley

Can you please explain to me the red part?

Why can it be written as neg, the 25 and 4?

Kid regards! Re: (x-a)^2 + (y-b)^2 = r^2 is the equation of a circle with radius r and   [#permalink] 15 Jun 2019, 10:50
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