Basic circles question (x^2 + y^2 -8x -2y -3 = 0)

[From: ] [author: ] [Date: 11-05-08] [Hit: ]

-3):x^2 + y^2 -8x -2y -3 = 6^2 + (-3)^2 - 8*6 + 2*3 - 3 = 0x - 2y - 12 = 6 + 2*3 - 12 = 0so this solution works!Thus, the point where the circle meets the line is (6, -3), which is the answer to part ii.-Good answer,......

x = 6

now we can use this value of x to solve for y in the first equation:

6^2 + y^2 -8*6 -2y -3 = 0
36 + y^2 - 48 - 2y - 3 = 0
y^2 - 2y - 15 = 0
(y - 5)(y + 3) = 0
so, we have that y = 5 or y = -3 when x = 6, giving two points of intersection for the line and the circle: (6, 5) and (6, -3)

Now, we should check that these solutions work:

for (6, 5):
x^2 + y^2 -8x -2y -3 = 6^2 + 5^2 - 8*6 - 2*5 - 3 = 0
x - 2y - 12 = 6 - 2*5 - 12 = -16
so this solution is NO good, because the second equation does not equal 0.

for (6, -3):
x^2 + y^2 -8x -2y -3 = 6^2 + (-3)^2 - 8*6 + 2*3 - 3 = 0
x - 2y - 12 = 6 + 2*3 - 12 = 0
so this solution works!

Thus, the point where the circle meets the line is (6, -3), which is the answer to part ii.

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Good answer, though you could shorten the last bit: once you know x = 6 is the only possible x value, the line can therefore only pass through one point on the circle, so just put this back into the line: y = x/2 - 6 = 6/2 - 6 = -3, so (6, -3) is the only point of intersection.

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Circle : x² + y² - 8x - 2y - 3 = 0

∴ ( x - 4 )² + ( y - 1 )² = (2√5)²

∴ Centre C ≡ ( 4, 1 ), Radius R = 2√5 ............. (1)
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[ i ]

If M is the mid-point of chord ST,

then : CM ⊥ ST at M.

∴ in triangle CMS :

. . CM = 2 (given), CS = R = 2√5, m∠ CMS = 90°

∴ by Pythagoras' Thm : (CM)² + (MS)² = (CS)²

∴ (MS)² = (CS)² - (CM)² = R² - 2² = 20 - 2² = 16

∴ MS = 4

∴ ST = 2( MS ) = 2( 4 ) = 8 .................... Ans.
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Happy To Help !
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