A circle in the xy-plane is defined by the equation (x - 3)^2 + (y - 2)^2 = 13. The line y… — SAT

A circle in the $xy$ -plane is defined by the equation $(x - 3)^2 + (y - 2)^2 = 13$ . The line $y = 3x$ intersects the circle at the origin $(0, 0)$ and at a second point $P$ . What is the $y$ -coordinate of point $P$ ?

A
$\frac{9}{5}$
B
$\frac{11}{5}$
$\frac{27}{5}$ Cevap
D
$-\frac{9}{5}$

Cevap

The y-coordinate of point P is

\frac{27}{5}

Substituting the linear relationship

y = 3x

into the circle equation yields a quadratic equation in terms of

x

. Solving this equation gives the

x

-coordinates of the intersection points as

x = 0

and

x = \frac{9}{5}

. Since point

P

is not the origin, its

x

-coordinate is

\frac{9}{5}

. Substituting this back into

y = 3x

gives the

y

-coordinate of point

P

\frac{27}{5}

Adım Adım Çözüm

Substitute the linear equation

y = 3x

into the equation of the circle.

(x - 3)^2 + (3x - 2)^2 = 13

Substitution reduces the system of two equations with two variables to a single quadratic equation with one variable.

Expand and simplify the equation.

10x^2 - 18x = 0

Expanding

(x - 3)^2

x^2 - 6x + 9

and

(3x - 2)^2

9x^2 - 12x + 4

, then combining like terms and subtracting 13 from both sides, isolates the quadratic terms.

Solve for

x

by factoring.

2x(5x - 9) = 0

, which gives

x = 0

x = \frac{9}{5}

The roots of the factored equation represent the x-coordinates of the two intersection points.

Find the y-coordinate of the second point

P

using

y = 3x

y = 3 \left(\frac{9}{5}\right) = \frac{27}{5}

Since point

P

is not the origin, its x-coordinate must be

\frac{9}{5}

. Substituting this into the linear equation gives its corresponding y-coordinate.

Anahtar Kavram

Solving nonlinear systems of equations involving a circle and a line by substitution.

Soru

Cevap

Adım Adım Çözüm

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