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Ad libitum [116K]
3 years ago
14

The endpoints of the longest chord on a circle are (4, 5.5) and (4, 10.5).the center of the circle is at the point , and its rad

ius is units. the equation of this circle in standard form is .
Mathematics
2 answers:
Fiesta28 [93]3 years ago
6 0

Answer:

The endpoints of the longest chord on a circle are (4, 5.5) and (4, 10.5).

The center of the circle is at the point  ( ? ), and its radius is (2.5) units. The equation of this circle in standard form is (x - 4)^2 + (y - 8)^2 = 6.25

i just did this and the last two were correct im not 100% sure the first one sorry

Step-by-step explanation:


Furkat [3]3 years ago
4 0
 The midpoint of the points ( 4, 5.5 ) and ( 4, 10.5 ) is the center of the circle:
 C =(\frac{4+4}{2}, \frac{5.5+10.5}{2})
 C = ( 4, 8 ),  h=4, k=8
 r =\sqrt{(4-4)^{2} +(8-5.5)^{2} } = \sqrt{0+2.5^{2} }
 r = 2.5
 The equation of the circle in standard form:
 ( x - h )² + ( y - k )² = r²
 ( x - 4 )² + ( y - 8 ) = 6.25    
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x = \frac{22}{7}

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Step-by-step explanation:

2x-3y=5\\5x=4y=14\\\\\left[\begin{array}{cc}2&-3\\5&-4\\\end{array}\right] \left[\begin{array}{c}x\\y\\\end{array}\right] =\left[\begin{array}{c}5\\14\\\end{array}\right]

Let A = \\\left[\begin{array}{cc}2&-3\\5&-4\\\end{array}\right]

The inverse of A multiplied by A = the identity matrix \left[\begin{array}{cc}1&0\\0&1\\\end{array}\right]

Inverse of A = \frac{1}{detA} \left[\begin{array}{ccc}-4&3\\-5&2\\\end{array}\right]

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\left[\begin{array}{ccc}x\\y\\\end{array}\right] = \left[\begin{array}{cc}\frac{-4}{7} &\frac{3}{7} \\\frac{-5}{7} &\frac{2}{7}\\\end{array}\right] \left[\begin{array}{ccc}5\\14\\\end{array}\right] = \left[\begin{array}{ccc}\frac{22}{7} \\\frac{3}{7} \\\end{array}\right]

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