+129852 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
+1252 \(\sqrt{34}\)
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!
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