In the standard (x, y) coordinate plane, a circle has the equation x^2 + y^2 + 10x - 4y + … — ACT

In the standard $(x, y)$ coordinate plane, a circle has the equation $x^2 + y^2 + 10x - 4y + 13 = 0$ . A line with a slope of $-\frac{1}{2}$ passes through the center of this circle. What is the $y$ -intercept of this line?

A
$-\frac{9}{2}$
B
$-\frac{7}{2}$
$-\frac{1}{2}$ Cevap
D
$\frac{1}{2}$
E
$\frac{9}{2}$

Cevap

The

y

-intercept of the line is

-\frac{1}{2}

Completing the square for the circle equation yields

(x + 5)^2 + (y - 2)^2 = 16

, which places the center at

(-5, 2)

. Substituting the center

(-5, 2)

and slope

m = -\frac{1}{2}

into the point-slope form

y - 2 = -\frac{1}{2}(x + 5)

simplifies to

y = -\frac{1}{2}x - \frac{1}{2}

, making the

y

-intercept

-\frac{1}{2}

Adım Adım Çözüm

Group terms and complete the square for both

x

and

y

in the circle equation.

(x^2 + 10x + 25) + (y^2 - 4y + 4) = -13 + 25 + 4

, which simplifies to

(x + 5)^2 + (y - 2)^2 = 16

Converting the general equation of a circle into standard form

(x - h)^2 + (y - k)^2 = r^2

reveals its center

(h, k)

Identify the center of the circle from the standard form equation.

The center is

(h, k) = (-5, 2)

Comparing

(x + 5)^2 + (y - 2)^2 = 16

(x - h)^2 + (y - k)^2 = r^2

gives

h = -5

and

k = 2

Use the point-slope formula

y - y_1 = m(x - x_1)

with point

(-5, 2)

and slope

m = -\frac{1}{2}

to find the equation of the line.

y - 2 = -\frac{1}{2}(x - (-5)) \implies y - 2 = -\frac{1}{2}x - \frac{5}{2} \implies y = -\frac{1}{2}x - \frac{1}{2}

Finding the slope-intercept form

y = mx + b

allows direct identification of the

y

-intercept

b

Anahtar Kavram

Converting circle equations to standard form to find center coordinates and finding linear equations from slope and point

Tahmini Süre:1m 15s

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Cevap

Adım Adım Çözüm

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