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1 year ago

the center of a circle is on the line y=2x and the line x=1 is tangent to the circle at (1,6).find the center and the radius

Mathematics

1 answer:

Julli [10]1 year ago

6 0

The radius and the center of the circle are 4 units and (1,2), respectively

<h3>How to determine the center and the radius?</h3>

The center of the circle is on

y = 2x and x = 1

Substitute x = 1 in y = 2x

y = 2 * 1

Evaluate

y = 2

This means that the center is

Center = (1, 2)

Also, we have the point of tangency to be:

(x, y) = (1, 6)

This point and the center have the same x-coordinate.

So, the distance between this point and the center is

d = 6 - 2

d = 4

This represents the radius

Hence, the radius and the center of the circle are 4 units and (1,2), respectively

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When will these two lines intersect <br>C. y=-2x +3<br>y = x

Andrej [43]

Answer:

The intersection point of the given lines is (1,1).

Step-by-step explanation:

Here, the given equations are:

y = - 2 x + 3

y = x

Substitute y = x in the first equation,

y = -2 x + 3 becomes y = -2 (y) + 3

or, y + 2 y = 3

or, 3 y = 3 ⇒ y = 3/3 = 1

⇒ y = 1

Now, as x = y ⇒ x = 1

Hence, the intersection points of the given lines is (1,1).

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3 years ago

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Solnce55 [7]

Answer:

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Step-by-step explanation:

∠ ACD and ∠ EFH are Alternate exterior angles and are congruent, thus

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x = 12

Thus

∠ EFH = 11x - 20 = 11(12) - 20 = 132 - 20 = 112°

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3 years ago

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Answer:

Step-by-step explanation:

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Please help and show all work so I can fully understand it, thanks.

seraphim [82]

Answer:

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Step-by-step explanation:

$\frac{\frac{4m-5}{m^4 -7m^3 +12m^2}}{\frac{4m-5}{m^3 -3m^2}}$

Factor the equation:

$\frac{\frac{4m-5}{m^2(m^2 -7m+12)}}{\frac{4m-5}{m^2(m-3)}}$

$\frac{\frac{4m-5}{m^2(m-3)(m-4)}}{\frac{4m-5}{m^2(m-3)}}$

Rewrite to suit the format of multiplying two fractions. Remember, dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second. A reciprocal of a fraction is when one switches the place of the numerator and the denominator, that is, the value on top (numerator), and the value on the bottom (denominator).

$\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}$

Simplify, take out common terms that are found on both the numerator and denominator

$\frac{4m-5}{m^2(m-3)(m-4)}*\frac{m^2(m-3)}{4m-5}$

$\frac{1}{m-4}$

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3 years ago