﻿ Tangent to Circle Worksheet | Problems & Solutions

# Tangent to Circle Worksheet

Tangent to Circle Worksheet
• Page 1
1.
$\stackrel{‾}{\mathrm{PN}}$, $\stackrel{‾}{\mathrm{PM}}$, and $\stackrel{‾}{\mathrm{AB}}$ are tangents to the circle as shown. If PM = 11 in., then find the perimeter of ΔPAB.

 a. 32 in. b. 27 in. c. 26 in. d. 22 in.

#### Solution:

Perimeter of Δ PAB = PA + AB + PB
[Sum of the sides.]

= PA + AQ + BQ + PB
[AB = AQ + QB.]

= PA + AN + BM + PB
[AQ = AN, BQ = BM, tangents from the same point to the circle are of equal length.]

= PN + PM = 11 + 11 = 22 in.
[Tangents from the same point to the circle are of equal length, PM is given as 11 in..]

2.
Find the perimeter of ΔABC, if AB = AC. The radius of the inscribed circle is 4 cm and a smaller circle of radius 3 cm is touching the incircle externally.

 a. $\frac{257}{\sqrt{12}}$ cm b. $\frac{256}{\sqrt{12}}$ cm c. $\frac{258}{\sqrt{12}}$ cm d. $\frac{255}{\sqrt{12}}$ cm

#### Solution:

Draw figure

From the figure, MD = ME = 4 cm

FN = NP = 3 cm

Let AP = x

AM / AN = EM / FN
[ΔAFN ~ ΔAEM.]

10+x3+x = 4 / 3
[Substitute.]

x = 18 cm
[Solve.]

AE = AM2-EM2 = 812

AD = AP + PD = 32 cm

Let BE = BD = y

(AE + y)2 = y2 + AD2 y = 1612cm

Perimeter of ΔABC = AB + BC + AC

AC = AB = AE + BE = 11212

BC = 2 × BD = 3212

Perimeter = 11212+3212+11212 = 25612 cm

3.
Find the radius of the circle inscribed in the square.

 a. 36 cm b. 24 cm c. 72 cm d. 30 cm

#### Solution:

Mark the points A, B, C, D, and E and specify the lengths.

Since the circle is inscribed in the square, the sides of the square form the tangents to the circle.
[A and E are the midpoints of the corresponding sides.]

BC = 12. So, CD = r - 12
[BD = OA = r.]

DE = 6. So, OD = r - 6
[OE = r.]

ODC is a right triangle. So, r2 = (r - 12)2 + (r - 6)2
[Use Pythagorean Theorem.]

r2 = (r2 - 24r + 144) + (r2 - 12r + 36)

r2 = 2r2 - 36r + 180

r2 - 36r + 180 = 0

(r - 30) (r - 6) = 0
[Factor.]

r = 30 cm
[Since 6 cm cannot be the radius.]

The radius of the circle is 30 cm.

4.
Name the line segment AB in the figure.

 a. Diameter b. Tangent c. Chord d. None of the above

#### Solution:

Any line that touches the circumference of a circle at a point is called the tangent.

In the figure, the line segment AB touches the circumference of the circle at the point D.

So, AB is called the tangent to the circle with center O.

5.
Distance between any two parallel tangents to a circle is equal to
 a. the diameter of the circle b. twice the diameter of the circle c. thrice the radius of the circle d. the radius of the circle

#### Solution:

From the above, distance between tangents = PQ = Diameter of the circle

6.
$\stackrel{‾}{\mathrm{PQ}}$ and $\stackrel{‾}{\mathrm{RS}}$ are the tangents to the circle with center C at P and R. CQ = 25 in., CS = 12 in.. Find the ratio of the lengths of the tangents at P and R after rounding to the nearest whole number. [Given CP = 9 in..]

 a. 26 : 15 b. 34 : 21 c. 16 : 3 d. 23 : 8

#### Solution:

CQ = 25 in., CS = 12 in..
[Given.]

Radius of the circle, CP = CR = 9 in..

Since ΔCPQ is right angled, CQ² = CP² + PQ² PQ² = CQ² - CP² PQ = 23.3 = 23 in..
[Rounded to the nearest whole number.]

Since ΔCRS is right angled, CS² = CR² + RS² RS² = CS² - CR² RS = 7.9 = 8 in..
[Rounded to the nearest whole number.]

The ratio of the lengths of the tangents at P and R is PQ: RS = 23 : 8

7.
$\stackrel{‾}{\mathrm{OL}}$ is perpendicular to $\stackrel{‾}{\mathrm{PL}}$, OL = 12 cm and OP = 31.2 cm. Find the length of $\stackrel{‾}{\mathrm{PL}}$.

 a. 28.8 cm. b. 48.8 cm. c. 43.8 cm. d. 38.8 cm.

#### Solution:

In right triangle OLP, PL = OP2-OL2
[Pythagorean theorem.]

PL = 31.22-122 = 28.8 cm.

8.
O is the center of the circle and PT = 11 in. Find PM.

 a. 11 in. b. less than 11 in. c. greater than 11 in. d. cannot be determined

#### Solution:

Tangents drawn from an external point to a circle are equal.

Therefore, PM = 11 in.

9.
If $a$ = 5 units, $b$ = 6 units, then find the value of $x$.

 a. 3.10 units b. 7.10 units c. 1.10 units d. 6.10 units

#### Solution:

ΔCPT is a right triangle.

CT2 = CP2 + PT2.

(x + 5)2 = x2 + 36

x = 1.10 units

10.
Graph the equation $x$² + ($y$ - 1)² = 9. Then draw a line segment from (6, 0) as a tangent to the circle. Find the length of this tangent.

 a. 4$\sqrt{2}$ units b. 2$\sqrt{7}$ units c. 2$\sqrt{3}$ units d. 1$\sqrt{2}$ units

#### Solution:

For x² + (y - 1)² = 9, center C (0, 1) and radius = 3 units.
[Given circle equation.]

PQ is the tangent to the circle C.

Draw CQ and CP.

ΔCQP is right triangle. Hence, CP² = CQ² + PQ²
[Apply pythagorean theorem to ΔCQP.]

But, CP = 37
[Use the formula for distance between two points.]

PQ² = CP² - CQ² = 37 - 9 = 28 PQ = 27 units
[CQ = radius = 3 units.]