point (2, k) lies on (x-3)^2 + (y-4)^2 = 16; find k

[2Advanced]

I've been trying to figure out how to solve this problem but the format is telling me to give my answer in Surds. I used the Quadratic Formula, but I'm still having no luck solving it. Here's the question:

The point (2, k) lies on the circle given by the equation (x - 3)^2+ (y - 4)^2 = 16

Find the possible values of k, giving your answers in the form a + b Sqrt(16).

Any help regarding how to get started and how to work through this would be appreciated. Many thanks!

 

[stapel]

It would have been helpful if you'd shown what you'd tried, so we could try to find where things are going awry. Lacking that, the usual solution process is as follows:

i) Plug the point (x, y) = (2, k) into the circle formula, since the point lies on the circle and thus must solve the equation.

ii) Simplify the (2 - 3)² bit of the equation, and subtract the value over to the right-hand side.

iii) Take the square root of both sides, remembering to put a "plus-minus" sign on the right-hand side.

iv) Add the "4" over to the right-hand side.

This will give you the solution in the form "a + bsqrt[15]". (I'm assuming the "16" is a typo, since that isn't the correct answer and since sqrt[16] = 4 anyway) The "plus-minus" gives you the two solutions. Note: The value of b is "1", in this case.

Eliz.

 

[2Advanced]

Thanks for the reply, Yeah, sorry about the typo. Your right its a+b sqrt(15)

My method was approach of making the k value as 0 to find the possible value what the missing y value would be. Other than that, my working at the end was kind of led to the answer: 1 ''plus-minus'' sign sqrt16

Explains why i didn't show my working out as it didn't look correct.

With the solution you've explained. I've managed to get my answer to a+b sqrt15 but how can i find the value of the k which im a bit confused?

Thanks for your help

 

[stapel]

I've managed to get my answer to a+b sqrt15 but how can i find the value of the k which im a bit confused?

I don't understand. The "a + bsqrt[15]" thing is the value of k. If you've followed the step-by-step instructions and gotten "k = a ± bsqrt[15]", how are you not finding the values of "k"?

Please reply showing all of your steps. Thank you.

Eliz.

 

[2Advanced]

Ah okies, heres my working out:

(x-3)^2+(y-4)^2=16
(2-3)^2 +(k-4)^2=16
(2-3)(2-3) +(k-4)^2=16
1+(k-4)^2=16
(k-4)^2=(+/-) sign sqrt15

= 4+1sqrt15

a+bsqrt15

Sorry about that, most likely im confused.

 

[2Advanced]

Heres another detailed solution to my working with a method what i've learnt. Hope its right.

(x-3)^2 +(y-4)^2=16

Step 1: (x-3)^2 => (x-3)(x-3) => x^2-3x-3x-9 => x^2-6x+9

Step 2: (y-4)^2 => (y-4)(y-4) => y^2-4y-4y-16 => y^2-8y-16

Substitute: 2 into x value in the equation: x^2-6x+9

2^2 - 6(2)+9
4-12+9
=1

1+y^2-8y+16=16

1+y^2-8y+1


Using the Quadratic Forumula:

-b(+/-)sign sqrtb^2-4ac over 2a

I plug in the values: a= 1, b=-8, c = 1 and multiply them together

=> 8(+/-) sign sqrt64-4 over 2

=> 8(+/-) sign sqrt60 over 2

=> 8(+/-) sign sqrt30

=> 4(+/-) 1 sqrt15


Value of a = 4, Value of b = 1

a+bsqrt15



Thanks for your help! Very much appreciated!