In the xy-coordinate system, a circle with radius root30 and center

Question

In the xy-coordinate system, a circle with radius  \sqrt {30}  and center (2,1) intersects the x-axis at (k,0). One possible value of k is.

A.  2 + \sqrt {26}

B.  2 + \sqrt {29}

C.  2 + \sqrt {31}

D.  2 + \sqrt {34}

E.  2 + \sqrt {35}

ENGRTOMBA2018 · Accepted Answer

Equation of circle in xy coordinates with center at (a,b) and radius r --->  (x-a)^2+(y-b)^2 = r^2

Per the question, a=2, b=1,  r^2 =30

We get the equation of the circle as  x^2+y^2-4x-2y = 25

Now the above equation should also be satisfied by the point (k,0) such that we get

k^2-4k-25 = 0  ----> k = 2+  \sqrt{29}  as 1 possible value. B is the correct answer.

nitika231 · Answer

How are you getting the highlighted portion as = 25? shouldn't it be 29? And in the last step, how did you eliminate the K from 4K?

ENGRTOMBA2018 · Answer

No, check your solution.

When we get,  (x-a)^2+(y-b)^2 = r^2  with a =2, b =1 and  r^2 =30, we get

(x-2)^2+(y-1)^2 = 30  ---->  x^2+4-4x+y^2+1-2y = 30 --->  x^2-4x+y^2-2y+5 = 30  ---->  x^2-4x+y^2-2y = 25 .

Thus, based on my post above, we get

k^2-4k-25 = 0

We know that for any quadratic equation,  px^2+qx+r =0  with p,q,r as constants, the solutions are :  x = \frac{-q \pm \sqrt{q^2-4pr}}{2p} , where p=1, q =-4, r=-25

Substituting the above values, we get, k = 2 + \sqrt 29

Hope this helps.

nitika231 · Answer

Understood! Thanks a lot for explaining it step-by-step! Maths has never been my strong suit, therefore, I struggle with what others may think is common knowledge. I realized I had miscalculated in one step and forgotten entirely about that quadratic formula in the second step. :oops:

Mistral · Answer

So we know the equation of the circle is  (x-h)^2 + (y-k)^2 = r^2 . Where (h,k) is the centre of the circle and r is the radius. 
In your question, we know (h,k) = (2,1) and r =  \sqrt{30}

Hence we get  (x-h)^2 + (y-k)^2 = r^2 as,
(x-2)^2 + (y-1)^2 = 30

now the point (k,0) should satisfy this circle's equation.

So lets replace x and y with k and 0 respectively.

(k-2)^2 + (0-1)^2 = 30

=> k^2 - 4k = 25

=> k^2 - 4k - 25 =0

=>  2 + \sqrt{29}

narendran1990 · Answer

Since the center is (2,1), drawing a perpendicular line to x-axis gives us a value (2,0). Given radius as root 30 & the point intersects x-axis at (k,0), therefore as per hypotenuse formula: 1^2+x^2=(root 30)^2. Solving we get x=root 29. Points of intersection is (root 29,0) & (2+root 29,0) Answer choice: B

dave13 · Answer

hey pushpitkc i encountered algebraic tech issues

how to solve it through factoring out k^2 - 4k - 25 =0 to get -25 the only numbers that get multiplied are -5*5 or -1*25 , which dont give me -4 when adding up Hey niks18 maybe you can help ?

pushpitkc is travelling I think

So what am I missing

niks18 · Answer

Hi  dave13

While factorization works well for most GMAT like questions but you should also remember the formula to calculate roots of the equation which you can use for complex quadratic equation such as the above.

The formula to calculate the roots of the quadratic equation is

-b±\sqrt{b^2-4ac}/2a

substitute the values of a,b & c in this formula, you will get your answer

dave13 · Answer

Hi niks18 thank you for the explanaton, i solved it through the formula that you provided yesterday

, but I couldnt solve it through factorization. can you please explain how to solve this k^2 - 4k - 25 =0 through factorozation

as i mentioned earlier, to get -25 the only numbers that get multiplied are -5*5 or -1*25 , which dont give me -4 when adding up

that is a stumbilng stone for me

so what am i doing wrong ?

thanks! have a great weekend

niks18 · Answer

Hi  dave13

Not all quadratic equation can be solved using factorization method. Hence there are multiple methods. this is one of the equation that cannot be solved using factorization.
consider another example  x^2-3x-3=0 . this equation can also be not solved using factorization. You need to use other methods depending upon the equation.

dkilroy · Answer

After using the quadratic formula, I am getting 2 +or- 2*\sqrt{29}

Abhibarua · Answer

The equation of the circle: (x-2)^2 + (y-1)^2 = 30

Since the circle intersects the x-axis, y = 0

(x-2)^2 + 1 = 30

(x-2)^2 = 29

Look at the options. The left-hand side and the right-hand side will be equal if x = 2+ square root over 29

Option: B

danbrown9445 · Answer

Theres no need to factorize/use quadartic formula.

Center of circle is (2,1).
Point on circle is (k,0)
Radius is  \sqrt {30}

(x-a)^2 + (y-b)^2 = r^2 
 (k-2)^2 + (-1)^2 = 30 
( k-2)^2 = 29

Now compare with answer choices by substituting value of k.
Its easy to see B is the correct choice.
 (2 + \sqrt {29} - 2)^2  = 29

bumpbot · Answer

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In the xy-coordinate system, a circle with radius root30 and center

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In the xy-coordinate system, a circle with radius root30 and center

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