Free Math Worksheets

- A radius is perpendicular to a tangent at the point of tangency. - Use the circle with diameter \(\overline{MQ}\) to locate the tangent points. - Apply the Pythagorean theorem to a triangle containing \(M\), \(Q\), and one tangent point. - Keep the radical for the exact value before rounding.

1. Draw circle \(k\), point \(Q\), and \(\overline{MQ}\). 2. Construct midpoint \(K\) of \(\overline{MQ}\), and draw the auxiliary circle centered at \(K\) with radius \(4\,\text{cm}\). Its diameter is \(\overline{MQ}\). 3. Label the intersections of the two circles \(T_1\) and \(T_2\). Draw \(\overleftrightarrow{QT_1}\) and \(\overleftrightarrow{QT_2}\). These lines are tangent because each radius is perpendicular to its line at the point of tangency. 4. In right triangle \(MT_1Q\), \(MT_1^2 + T_1Q^2 = MQ^2\). 5. Substitute: \(3^2 + T_1Q^2 = 8^2\), so \(T_1Q^2 = 64 - 9 = 55\). 6. Therefore, \(T_1Q = T_2Q = \sqrt{55}\,\text{cm} \approx 7.42\,\text{cm}\).

Each tangent segment has length \(\sqrt{55}\,\text{cm} \approx 7.42\,\text{cm}\).

- A tangent is perpendicular to the radius at the point of tangency. - An angle subtending a diameter is a right angle. - Classify \(Q\) as outside, on, or inside the circle.

1. Draw \(\overline{MP}\) and construct its midpoint \(Z\). 2. Draw the circle with diameter \(\overline{MP}\). Let its intersections with circle \(k\) be \(T_1\) and \(T_2\). 3. Draw \(\overleftrightarrow{PT_1}\) and \(\overleftrightarrow{PT_2}\). Because each angle subtending diameter \(\overline{MP}\) is a right angle, \(MT_1 \perp PT_1\) and \(MT_2 \perp PT_2\). A line perpendicular to a radius at its endpoint is tangent to the circle. 4. If \(MQ > 3\,\text{cm}\), then \(Q\) is outside the circle and there are two tangents. 5. If \(MQ = 3\,\text{cm}\), then \(Q\) is on the circle and there is one tangent, perpendicular to \(\overline{MQ}\) at \(Q\). 6. If \(MQ < 3\,\text{cm}\), then \(Q\) is inside the circle and there are no tangent lines through \(Q\).

a) Intersect circle \(k\) with the circle whose diameter is \(\overline{MP}\). Joining \(P\) to the two intersection points gives the tangents. b) \(MQ > 3\,\text{cm}\): two tangents. \(MQ = 3\,\text{cm}\): one tangent. \(MQ < 3\,\text{cm}\): no tangents.

- The diameter is twice the radius. - Identify the diameter of the auxiliary circle. - Compare the location of \(A\) with the interior, boundary, and exterior of the original circle.

1. Since \(\overline{MA}\) is the diameter of the auxiliary circle, \(MA = 2\cdot 5.5 = 11\,\text{cm}\). 2. Point \(T_1\) lies on the circle with diameter \(\overline{MA}\), so \(\angle MT_1A = 90^\circ\) by the inscribed-angle theorem for a diameter. 3. This right angle makes \(\overline{AT_1}\) perpendicular to radius \(\overline{MT_1}\), so \(\overline{AT_1}\) is tangent to the original circle. 4. No tangent can be drawn when \(A\) lies inside the original circle, meaning \(MA\) is less than that circle's radius. If \(A\) lies on the circle, there is exactly one tangent.

a) \(MA = 11\,\text{cm}\). b) \(\angle MT_1A = 90^\circ\), by the inscribed-angle theorem for a diameter. c) Tangent construction is impossible when \(A\) lies inside the original circle.

- The auxiliary circle's diameter is \(\overline{MP}\). - Its center is the midpoint of that diameter. - Imagine the tangent lines from a point very far from the circle.

1. Since \(P\) lies on the outer circle, \(MP = 6\,\text{cm}\). 2. The auxiliary circle has \(\overline{MP}\) as its diameter, so its radius is \(3\,\text{cm}\). 3. Point \(K\) is the midpoint of \(\overline{MP}\), so \(MK = 3\,\text{cm}\). 4. As \(P\) moves farther from \(M\), the tangent lines become closer to parallel. The tangent points move away from the ray \(MP\) toward the endpoints of the diameter perpendicular to \(\overline{MP}\). 5. Therefore, the tangent points move farther apart, and the minor arc between them increases toward a semicircle.

a) The auxiliary-circle radius is \(3\,\text{cm}\). b) \(MK = 3\,\text{cm}\). c) The tangent points move farther apart toward the endpoints of the diameter perpendicular to \(\overline{MP}\); the minor arc between them increases.

- Determine whether \(g\) is horizontal or vertical. - Recall the angle between a tangent and the radius to its point of tangency. - Lines perpendicular to the same line are parallel. - Move one radius from the center in both directions along \(g\).

1. Line \(g\) is the vertical line \(x = 5\). Tangents perpendicular to \(g\) must be horizontal. 2. A tangent is perpendicular to the radius at the point of tangency. Therefore, the radii to \(T_1\) and \(T_2\) must lie on \(g\). 3. Move \(3\) units up and down from \(M(5, 5)\): \(5 + 3 = 8\) and \(5 - 3 = 2\). 4. The points of tangency are \(T_1(5, 8)\) and \(T_2(5, 2)\). The tangent lines are \(y = 8\) and \(y = 2\).

1. The tangents are the horizontal lines \(y = 8\) and \(y = 2\). 2. \(T_1(5, 8)\) and \(T_2(5, 2)\)