Tangent to Circle Worksheet

**Page 1**

1.

$\stackrel{\u203e}{\mathrm{PN}}$, $\stackrel{\u203e}{\mathrm{PM}}$, and $\stackrel{\u203e}{\mathrm{AB}}$ 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. |

[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..]

Correct answer : (4)

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 |

Draw figure

From the figure, MD = ME = 4 cm

FN = NP = 3 cm

Let AP =

[ΔAFN ~ ΔAEM.]

[Substitute.]

[Solve.]

AE =

AD = AP + PD = 32 cm

[Substitute and add.]

Let BE = BD =

AB

(AE +

Perimeter of ΔABC = AB + BC + AC

AC = AB = AE + BE =

BC = 2 × BD =

Perimeter =

Correct answer : (2)

3.

Find the radius of the circle inscribed in the square.

a. | 36 cm | ||

b. | 24 cm | ||

c. | 72 cm | ||

d. | 30 cm |

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 =

[BD = OA =

DE = 6. So, OD =

[OE =

ODC is a right triangle. So,

[Use Pythagorean Theorem.]

(

[Factor.]

[Since 6 cm cannot be the radius.]

The radius of the circle is 30 cm.

Correct answer : (4)

4.

Name the line segment AB in the figure.

a. | Diameter | ||

b. | Tangent | ||

c. | Chord | ||

d. | None of the above |

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.

Correct answer : (2)

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 |

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

Correct answer : (1)

6.

$\stackrel{\u203e}{\mathrm{PQ}}$ and $\stackrel{\u203e}{\mathrm{RS}}$ 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 |

[Given.]

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

Since ΔCPQ is right angled, CQ² = CP² + PQ²

[Rounded to the nearest whole number.]

Since ΔCRS is right angled, CS² = CR² + RS²

[Rounded to the nearest whole number.]

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

Correct answer : (4)

7.

$\stackrel{\u203e}{\mathrm{OL}}$ is perpendicular to $\stackrel{\u203e}{\mathrm{PL}}$, OL = 12 cm and OP = 31.2 cm . Find the length of $\stackrel{\u203e}{\mathrm{PL}}$.

a. | 28.8 cm. | ||

b. | 48.8 cm. | ||

c. | 43.8 cm. | ||

d. | 38.8 cm. |

[Pythagorean theorem.]

PL =

Correct answer : (1)

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 |

Therefore, PM = 11 in.

Correct answer : (1)

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 |

CT

(

Correct answer : (3)

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 |

[Given circle equation.]

Draw

ΔCQP is right triangle. Hence, CP² = CQ² + PQ²

[Apply pythagorean theorem to ΔCQP.]

But, CP =

[Use the formula for distance between two points.]

PQ² = CP² - CQ² = 37 - 9 = 28

[CQ = radius = 3 units.]

Correct answer : (2)