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3 years ago

Write the equation for the line tangent to the circle x2 + y2 = 29 at the point (2, 5).

1 answer:

3 0

X2 + y2 = 29

Center of the circle
x=0 and y=0

Therefore the slope of the line perpendicular to the tangent line
(y2-y1) / (x2-x1)
(5-0) / (2-0)
5/2

so the slope of the tangent
-2/5

General equation for the line
y = mx + c
if the line satisfies with the point (2,5)
5 = 2m + c

m = -2/5

5 = 2 * (-2/5) + c
c = 5 + (4/5)
c = 29/5

Therefore the equation of the line
y = mx + c
y = (-2/5) m + 29/5

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Write the equation of the line that passes through the points (-4,8) and (-2, -1).

IRINA_888 [86]

Answer:

y=-9/2x-10

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-1-8)/(-2-(-4))

m=-9/(-2+4)

m=-9/2

y-y1=m(x-x1)

y-8=-9/2(x-(-4))

y-8=-9/2(x+4)

y=-9/2x-36/2+8

y=-9/2x-18+8

y=-9/2x-10

5 0

2 years ago

Is the solution of this equation extraneous?

UkoKoshka [18]

Answer:

x = - 3 is extraneous

Step-by-step explanation:

Given

$\sqrt{x+7}$ - 1 = x ( add 1 to both sides )

$\sqrt{x+7}$ = x + 1 ( square both sides )

x + 7 = (x + 1)² ← distribute right side

x + 7 = x² + 2x + 1 ( subtract x + 7 from both sides )

0 = x² + x - 6 ← in standard form

0 = (x + 3)(x - 2) ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 2 = 0 ⇒ x = 2

As a check

Substitute these values into the equation and if both sides are equal then they are the solutions.

x = - 3 : $\sqrt{-3+7}$ - 1 = $\sqrt{4}$ - 1 = 2 - 1 = 1 ≠ - 3

x = 2 : $\sqrt{2+7}$ - 1 = $\sqrt{9}$ - 1 = 3 - 1 = 2

x = 2 is a solution and x = - 3 is extraneous

6 0

2 years ago

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Fiesta28 [93]

Answer:

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Step-by-step explanation:

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2 years ago

Find the center of the circle that you can circumscribe about DABC.

Vlada [557]

The center of the circle that circumscribe about DABC is $(h,k) = (1, 5)$ .

A circle can be modelled after a expression of the form:

$A\cdot x + B\cdot y + C =-x^{2}-y^{2}$ (1)

We can determine all coefficients by knowing three <em>distinct</em> points on plane.

If we know that $(x_{1}, y_{1}) = (2, 8)$ , $(x_{2}, y_{2}) = (0, 8)$ and $(x_{3}, y_{3}) = (2, 2)$ , then the solution of the system of linear equations is:

$2\cdot A + 8\cdot B + C = -68$ (2)

$8\cdot B + C = -64$ (3)

$2\cdot A + 2\cdot B + C = -8$ (4)

$A = -2, B = -10, C = 16$

Now we proceed to <em>complete</em> squares and factor each resulting perfect square trinomial in order to determine the coordinates of the center of the circle:

$x^{2}+y^{2}-2\cdot x -10\cdot y +16 = 0$