In the xy-plane, a circle with radius r, where r > 1, has its center in the first quadrant… — SAT

In the $xy$ -plane, a circle with radius $r$ , where $r > 1$ , has its center in the first quadrant. The circle is tangent to the line $x = 1$ and tangent to the line $y = 2$ . If the center of the circle lies on the line with equation $y = \frac{4}{3}x$ , what is the value of $r$ ?

Cevap: 2

Cevap

The radius

r

of the circle is 2.

Since the circle has radius

r

and is tangent to the lines

x = 1

and

y = 2

, its center

(h, k)

lies at a distance of

r

from both lines. This means

h = 1 \pm r

and

k = 2 \pm r

. Because the center is in the first quadrant,

h > 0

and

k > 0

. Given

r > 1

, the choice

h = 1 - r

would make

h

negative, so we must have

h = r + 1

. If

k = 2 - r

, the center is

(r + 1, 2 - r)

, and substituting this into the line

y = \frac{4}{3}x

gives

2 - r = \frac{4}{3}(r + 1) \implies 6 - 3r = 4r + 4 \implies 7r = 2 \implies r = \frac{2}{7}

, which contradicts the condition that

r > 1

. Therefore, we must have

k = r + 2

. Substituting the center

(r + 1, r + 2)

into

y = \frac{4}{3}x

yields

r + 2 = \frac{4}{3}(r + 1) \implies 3(r + 2) = 4(r + 1) \implies 3r + 6 = 4r + 4 \implies r = 2

Adım Adım Çözüm

Set up equations for the center coordinates

(h, k)

in terms of the radius

r

|h - 1| = r

and

|k - 2| = r

The distance from the center of a circle to any of its tangent lines is equal to the radius

r

Determine the correct sign for the absolute value expression of the

x

-coordinate.

h = r + 1

Since the center lies in the first quadrant,

h

must be positive. If

h = 1 - r

, then

r > 1

would imply

h < 0

, which is a contradiction.

Determine the correct sign for the absolute value expression of the

y

-coordinate by evaluating both possibilities on the line

y = \frac{4}{3}x

k = r + 2

k = 2 - r

, then substituting into the line equation gives

r = \frac{2}{7}

, which contradicts the condition that

r > 1

. Thus,

k

must equal

r + 2

Substitute

(r + 1, r + 2)

into the line equation

y = \frac{4}{3}x

and solve for

r

r = 2

Substituting gives

r + 2 = \frac{4}{3}(r + 1) \implies 3(r + 2) = 4(r + 1) \implies 3r + 6 = 4r + 4 \implies r = 2

Anahtar Kavram

Equations of Circles in the Coordinate Plane

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