A circle's center is at $(2, 4)$ $(2 ,4 )$ and it passes through $(7, 6)$ $(7 ,6 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?

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A circle's center is at $(2, 4)$ $(2 ,4 )$ and it passes through $(7, 6)$ $(7 ,6 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?

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Answer 1 · 2016-11-04T13:03:11

Length of the arc: $31.7$ $31.7$ (approx.)

Explanation:

If the circle has a center at $(2, 4)$ $(2,4)$ and passes through $(7, 6)$ $(7,6)$ then it has a radius of
$XXX r = \sqrt{{(7 - 2)}^{2} + {(6 - 4)}^{2}} = \sqrt{25 + 4} = \sqrt{29}$ $color(white)("XXX")r=sqrt((7-2)^2+(6-4)^2)=sqrt(25+4) =sqrt(29)$
and a diameter of
$XXX d = 2 r = 2 \sqrt{29}$ $color(white)("XXX")d=2r=2sqrt(29)$

The circumference of the circle would be
$XXX {Circumference}_{\circ} = π d = 2 \sqrt{29} π$ $color(white)("XXX")"Circumference"_circ=pid = 2sqrt(29)pi$

The complete circle's circumference is covered by an arc of $2 π = \frac{16 π}{8}$ $2pi=(16pi)/8$

An arc of $\frac{15 π}{8}$ $(15pi)/8$ will cover $\frac{\frac{15 π}{8}}{\frac{16 π}{8}} = \frac{15}{16}$ $((15pi)/8)/((16pi)/8)=15/16$ of ${Circumference}_{\circ}$ $"Circumference"_circ$

The given arc will cover
$XXX \frac{15}{16} \times 2 \sqrt{29} π$ $color(white)("XXX")15/16xx2sqrt(29)pi$

$XXX \approx 31.72123912$ $color(white)("XXX")~~31.72123912$ (with a calculator)

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A circle's center is at $(2, 4)$ $(2 ,4 )$ and it passes through $(7, 6)$ $(7 ,6 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?

1 Answer

Explanation:

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A circle's center is at (2 ,4 )(2,4)(2 ,4 ) and it passes through (7 ,6 )(7,6)(7 ,6 ). What is the length of an arc covering (15pi ) /8 15π8(15pi ) /8 radians on the circle?

1 Answer

Explanation:

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A circle's center is at $(2, 4)$ $(2 ,4 )$ and it passes through $(7, 6)$ $(7 ,6 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?