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# Circles

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3 please help with question C

thanks

Nov 7, 2018

### 3+0 Answers

#1
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a)   (x - 2)^2  + (y - 1)^2   = 13        (1)

Second part.... expand (1)   and we have

x^2 - 4x + 4    + y^2 - 2y + 1  = 13     simplify

x^2 + y^2 - 4x - 2y  +  5   = 13       subtract 13 from both sides

x^2 + y^2 -4x - 2y - 8  =   0

b)  A  = (4, k)

Sub 4 into (1)  for x

(4 - 2)^2  + ( y - 1)^2  = 13

2^2  + (y - 1)^2  = 13

4 + (  y - 1)^2  = 13

(y - 1)^2  = 9       take both roots

y - 1  = ±3

y =  ±3 + 1

So  y  =  3 + 1  = 4     or  y = -3 + 1  =  -2

So..... k  =  4     or k = -2

c) The tangent will meet the circle's radius at a  right angle...so...we have the right triangle ABC

AC will be one leg of the triangle  = sqrt (13)

BC will form the hypotenuse  = 7

And AB will form the other leg

So....using the Pythagorean Theorem

AB  =  sqrt  [  7^2  -  (sqrt (13) )^2  ]   = sqrt [ 49 - 13 ] = sqrt (36)  =  6   Nov 7, 2018
#3
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thank you very much!

YEEEEEET  Nov 7, 2018
#2
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$$\sqrt{34}$$ Use the Pythagorean Theorem for it, $$BC$$ is 7, and if we put a point $$D$$ at $$(4, 1)$$,

$$DA = 3, DC = BC-DC = 7-2=5$$

The right angle is ADB, and $$AD^2+DB^2=AB^2, 3^2+5^2=AB^2, 34=AB^2, \sqrt{34}=AB$$

Therefore, $$AB = \sqrt{34}$$

You are very welcome!

:P

Nov 7, 2018