In the xy-plane, a system of equations consists of the circle with equation (x - 7)^2 + (y… — SAT

In the $xy$ -plane, a system of equations consists of the circle with equation $(x - 7)^2 + (y - 4)^2 = 1$ and the line $y = kx$ , where $k$ is a positive constant. If the system has exactly one real solution and $k > 0.5$ , what is the value of $k$ ?

Answer: 0.75

Answer

The correct answer is 3/4 (or 0.75).

The correct answer is

3/4

(or

0.75

). Substituting the line

y = kx

into the circle equation and setting the discriminant of the resulting quadratic equation to zero yields a quadratic in

k

48k^2 - 56k + 15 = 0

. Solving this equation gives two positive tangent slopes:

k = 5/12

and

k = 3/4

. Since the problem specifies that

k > 0.5

, we choose

k = 3/4

, which equals

0.75

. Alternatively, using geometry, the distance from the center

(7,4)

to the line

kx - y = 0

must equal the radius

1

. This gives

\frac{|7k-4|}{\sqrt{k^2+1}} = 1

. Squaring both sides yields

(7k-4)^2 = k^2+1

, which simplifies to

48k^2 - 56k + 15 = 0

, giving the same values of

k

Step-by-Step Solution

Substitute the linear equation

y = kx

into the circle's quadratic equation

(x - 7)^2 + (y - 4)^2 = 1

(x - 7)^2 + (kx - 4)^2 = 1

To find the points of intersection between the line and the circle, substitute the expression for

y

from the linear equation into the circle's equation.

Expand the terms and write the equation in the standard quadratic form

Ax^2 + Bx + C = 0

(1 + k^2)x^2 - (14 + 8k)x + 64 = 0

Expanding

(x - 7)^2

yields

x^2 - 14x + 49

and expanding

(kx - 4)^2

yields

k^2x^2 - 8kx + 16

. Grouping the terms by powers of

x

and subtracting

1

from both sides gives the standard form.

Set the discriminant of the quadratic equation to

0

to satisfy the condition of exactly one real solution.

\Delta = [-(14 + 8k)]^2 - 4(1 + k^2)(64) = 0

A quadratic equation has exactly one real solution if and only if its discriminant,

B^2 - 4AC

, is equal to zero. Geometrically, this means the line is tangent to the circle.

Simplify the discriminant equation and solve for

k

48k^2 - 56k + 15 = 0

, which factors as

(12k - 5)(4k - 3) = 0

. Thus,

k = \frac{5}{12}

k = \frac{3}{4}

Simplifying

(14 + 8k)^2 - 256(1 + k^2) = 0

gives

196 + 224k + 64k^2 - 256 - 256k^2 = 0

, which simplifies to

-192k^2 + 224k - 60 = 0

. Dividing the entire equation by

-4

yields the quadratic equation in

k

Select the value of

k

that satisfies the given condition

k > 0.5

k = \frac{3}{4}

(or

0.75

)

Since

\frac{5}{12} \approx 0.417

is less than

0.5

and

\frac{3}{4} = 0.75

is greater than

0.5

, the correct value is

3/4

Key Concept

Solving a nonlinear system of equations involving a circle and a line by setting the discriminant of the substituted quadratic equation to zero to find the slope of the tangent lines.

Estimated Time:2m 30s

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