A circle in the xy-plane is defined by the equation (x - h)^2 + (y - k)^2 = 16, where h an… — SAT

A circle in the $xy$ -plane is defined by the equation $(x - h)^2 + (y - k)^2 = 16$ , where $h$ and $k$ are constants. The center $(h, k)$ of the circle lies on the line $y = x$ . A second line, which passes through the origin and has a slope of $-\frac{3}{4}$ , is tangent to the circle at exactly one point $(x, y)$ . If this point of tangency lies in a quadrant where $x > 0$ and $y < 0$ , what is the value of $h$ ?

A
$20$
B
$-\frac{20}{7}$
$\frac{20}{7}$ Cevap
D
$\frac{80}{7}$

Cevap

\frac{20}{7}

The correct option is

\frac{20}{7}

. By setting the center of the circle to

(h, h)

and the equation of the line to

3x + 4y = 0

, we find that the distance from the center to the line is

\frac{|7h|}{5}

. Since the line is tangent to the circle, this distance must equal the radius, which is

4

. This gives

h = \pm\frac{20}{7}

. Finding the point of tangency shows that

x = \frac{4}{25}h

and

y = -\frac{3}{25}h

. For the point of tangency to lie in Quadrant IV (

x > 0

and

y < 0

h

must be positive, which yields

h = \frac{20}{7}

Adım Adım Çözüm

Express the center of the circle and the equation of the tangent line in terms of the given parameters.

Since the center

(h, k)

lies on the line

y = x

, we have

k = h

. The circle has radius

R = \sqrt{16} = 4

and is centered at

(h, h)

. The tangent line passes through the origin with slope

-\frac{3}{4}

, so its equation is

y = -\frac{3}{4}x

, which simplifies to

3x + 4y = 0

Setting up the algebraic expressions for both geometric entities is necessary to relate them using coordinate geometry formulas.

Apply the tangency condition using the point-to-line distance formula.

The distance from the center

(h, h)

to the line

3x + 4y = 0

must equal the radius

4

. Thus:

\frac{|3h + 4h|}{\sqrt{3^2 + 4^2}} = 4 \implies \frac{|7h|}{5} = 4 \implies |7h| = 20

A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line equals the radius.

Solve for the possible values of

h

h = \pm\frac{20}{7}

Solving the absolute value equation yields two symmetric possibilities for the x-coordinate of the circle's center.

Determine the relationship between the center

h

and the coordinates of the point of tangency

(x, y)

to apply the quadrant constraint.

The radius connecting the center

(h, h)

to the point of tangency

(x, y)

is perpendicular to the tangent line. Since the tangent line has a slope of

-\frac{3}{4}

, the perpendicular radius line has a slope of

\frac{4}{3}

. Its equation is:

y - h = \frac{4}{3}(x - h) \implies y = \frac{4}{3}x - \frac{1}{3}h

The intersection of the perpendicular radius line and the tangent line will locate the exact point of tangency.

Solve the system of equations for the point of tangency

(x, y)

in terms of

h

Equating the tangent line and the perpendicular line:

-\frac{3}{4}x = \frac{4}{3}x - \frac{1}{3}h \implies -\frac{9}{12}x = \frac{16}{12}x - \frac{4}{12}h \implies -\frac{25}{12}x = -\frac{4}{12}h \implies x = \frac{4}{25}h

. Substituting back:

y = -\frac{3}{4}\left(\frac{4}{25}h\right) = -\frac{3}{25}h

This yields the coordinates of the tangency point as a function of the parameter

h

Apply the quadrant constraint (

x > 0

and

y < 0

) to choose the correct sign of

h

We require

x = \frac{4}{25}h > 0

and

y = -\frac{3}{25}h < 0

. Both inequalities are satisfied if and only if

h > 0

. Therefore,

h = \frac{20}{7}

This filters out the extraneous geometric solution that lies in Quadrant II.

Anahtar Kavram

Solving systems of nonlinear equations representing circles and lines by utilizing geometric relations, distance formulas, and quadrant constraints.

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