Free math worksheets

- A radius is perpendicular to a tangent at the point of tangency. - What auxiliary circle would force a right angle at each desired tangency point? - After the construction, identify a right 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, \(T_1Q=T_2Q=\sqrt{55}\,\text{cm}\approx7.42\,\text{cm}\).

Each tangent segment has length \(\sqrt{55}\,\text{cm}\approx7.42\,\text{cm}\).

- A tangent is perpendicular to the radius at the point of tangency. - What auxiliary circle would make an angle at a desired tangency point a right angle? - For part b, classify \(Q\) as outside, on, or inside the original 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 \(T_2(5, 2)\)

- Begin with the segment joining the center to the point on the circle. - Consider the distance from the center to another point on the constructed line. - Use the right triangle formed by the center, the tangency point, and that other point.

1. Line \(t\) passes through \(T\) and is perpendicular to radius \(\overline{OT}\). 2. For any other point \(Q\) on \(t\), triangle \(OTQ\) is right at \(T\), so \(OQ^2=OT^2+TQ^2\). 3. Since \(Q\ne T\), \(TQ^2>0\), and therefore \(OQ>OT\). Thus every other point on \(t\) lies outside the circle. 4. Hence \(t\) meets the circle only at \(T\) and is the tangent at \(T\).

Construct the line through \(T\) perpendicular to radius \(\overline{OT}\). Every other point \(Q\) on that line satisfies \(OQ>OT\), so it lies outside the circle. Therefore, the constructed line is tangent at \(T\).

- Construct each tangent using its radius at the tangency point. - Identify two points that are each equidistant from \(A\) and \(B\). - Two distinct points determine the perpendicular-bisector line.

1. The app draws the line through \(A\) perpendicular to \(OA\) and the line through \(B\) perpendicular to \(OB\). These are the tangents at \(A\) and \(B\), and they meet at \(P\). 2. Tangent segments from the same external point are congruent, so \(PA=PB\). Therefore, \(P\) lies on the perpendicular bisector of \(\overline{AB}\). 3. Radii \(OA\) and \(OB\) are congruent, so \(O\) is also equidistant from \(A\) and \(B\). Therefore, \(O\) lies on the same perpendicular bisector. 4. Since the perpendicular bisector is the unique line through the two distinct points \(O\) and \(P\), line \(OP\) is the perpendicular bisector of \(\overline{AB}\).

The tangents are perpendicular to \(OA\) and \(OB\). Because \(PA=PB\) and \(OA=OB\), both \(P\) and \(O\) lie on the perpendicular bisector of \(AB\). Hence \(OP\) is that perpendicular bisector.

- Tangency at a specified point determines a line on which the center must lie. - Tangency to both intersecting lines requires equal perpendicular distances from the center. - Combine the two center loci and count their intersections.

1. Construct the line \(n\) through \(T\) perpendicular to \(\ell\). The center of any circle tangent to \(\ell\) at \(T\) must lie on \(n\). 2. Construct the internal and external angle-bisector lines of \(\ell\) and \(m\). A point on either bisector is equidistant from the two lines. 3. Let \(C_1\) and \(C_2\) be the intersections of \(n\) with the two angle-bisector lines. 4. Draw the circle centered at \(C_i\) with radius \(C_iT\), for \(i=1,2\). Because \(C_iT\perp\ell\), each circle is tangent to \(\ell\) at \(T\). 5. Since each \(C_i\) lies on an angle bisector, its perpendicular distance to \(m\) equals its distance \(C_iT\) to \(\ell\). Thus each circle is also tangent to \(m\). 6. Any valid center must lie on both \(n\) and one of the two angle-bisector lines. These intersections are exactly \(C_1\) and \(C_2\), so there are exactly two circles.

Intersect the perpendicular to \(\ell\) at \(T\) with the internal and external angle-bisector lines of \(\ell\) and \(m\). The two intersection points are the centers; use each center's distance to \(T\) as its radius.

- Look for a construction that forces a right angle at a point on the given circle. - Use \(\overline{OP}\) as the diameter of an auxiliary circle. - Relate a right angle between a radius and a segment to tangency.

1. Construct the midpoint \(M\) of \(\overline{OP}\). 2. Draw the circle centered at \(M\) with radius \(MO\). This circle has \(\overline{OP}\) as a diameter. 3. Let the two intersections of this auxiliary circle with the original circle be \(T_1\) and \(T_2\). 4. Because \(\overline{OP}\) is a diameter of the auxiliary circle, \(\angle OT_1P\) and \(\angle OT_2P\) are right angles. 5. Thus \(OT_1\perp PT_1\) and \(OT_2\perp PT_2\). A line perpendicular to a radius at its endpoint on the circle is tangent, so \(\overline{PT_1}\) and \(\overline{PT_2}\) are the two tangent segments.

Construct the circle with diameter \(\overline{OP}\). Its intersections \(T_1\) and \(T_2\) with the given circle determine the tangent segments \(\overline{PT_1}\) and \(\overline{PT_2}\), because each forms a right angle with the corresponding radius.

- Replace the two unequal radii with their difference. - First find a direction tangent from one center to the reduced circle. - How can a parallel offset restore the smaller circle's radius while keeping the tangent direction?

1. Draw the auxiliary circle centered at \(O_1\) with radius \(5-2=3\,\text{cm}\). 2. Construct the two tangents from \(O_2\) to the auxiliary circle. Let one tangent touch it at \(T\). 3. Radius \(O_1T\) is perpendicular to tangent \(O_2T\). 4. On ray \(O_1T\), mark point \(V\) so that \(O_1V=5\,\text{cm}\). Through \(V\), construct line \(n\parallel O_2T\). 5. Drop the perpendicular from \(O_2\) to \(n\), meeting \(n\) at \(U\). Since the distance between the parallel lines \(O_2T\) and \(n\) is \(O_1V-O_1T=5-3=2\,\text{cm}\), \(O_2U=2\,\text{cm}\). 6. Line \(n\) is perpendicular to both radii \(O_1V\) and \(O_2U\), so it is tangent to both original circles. 7. Repeating the construction with the other auxiliary tangent gives the second common external tangent.

Shrinking the larger radius by the smaller radius reduces the problem to tangents from \(O_2\) to a circle of radius \(3\,\text{cm}\). Parallel offsets of those two tangent directions produce the two common external tangents.