I'm having trouble with this maths question. Please help!
Find the length cut off on the x-axis and y-axis by the circle x^2+y^2-2x-4y = 20
Find the length cut off on the x-axis and y-axis by the circle x^2+y^2-2x-4y = 20
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Take the x axis first. The circle intersects the x axis when y=0, and the x values satisfy:
x² + 0² - 2x - 4(0) = 20 ... after setting y=0; then simplify
x² - 2x = 20 ... add 1 to both sides to complete the square
x² - 2x + 1 = 21 ... left side factors now
(x - 1)² = 21 ... this solves easily
x = 1 ± √21
That's √21 away from x=1 on eitther side, so the length is 2√21
Do the same for the y axis where x=0:
0² + y² - 2(0) - 4y = 20
y² - 4y = 20
y² - 4y + 4 = 24
(y - 2)² = 24
y = 2 ± √24 = 2 ± 2√6
...and the length is 4√6
x² + 0² - 2x - 4(0) = 20 ... after setting y=0; then simplify
x² - 2x = 20 ... add 1 to both sides to complete the square
x² - 2x + 1 = 21 ... left side factors now
(x - 1)² = 21 ... this solves easily
x = 1 ± √21
That's √21 away from x=1 on eitther side, so the length is 2√21
Do the same for the y axis where x=0:
0² + y² - 2(0) - 4y = 20
y² - 4y = 20
y² - 4y + 4 = 24
(y - 2)² = 24
y = 2 ± √24 = 2 ± 2√6
...and the length is 4√6
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Solve the eqn for x=0 to get the y intercept and likewise
for the y=0 x intercept. From there on your on your own..
for the y=0 x intercept. From there on your on your own..