Circles

Solution

A tangent to a circle is a line that intersects the circle at exactly one point, known as the point of contact. The key property of a tangent is that it is perpendicular to the radius at the point of contact. This means that if you draw a radius to the point where the tangent touches the circle, the tangent will form a right angle (90 degrees) with the radius. This property is crucial for solving various problems related to circles, such as finding the length of a tangent from an external point or proving certain geometric theorems. For example, in real life, the wheels of a bicycle are tangent to the road at the point of contact, and the radius of the wheel is perpendicular to the road at that point.

Solution

To prove that the lengths of two tangents drawn from an external point to a circle are equal, consider an external point P outside the circle. Draw two tangents PQ and PR from P to the circle, touching it at points Q and R respectively. Join OQ, OR, and OP. Since PQ and PR are tangents, ∠OQP and ∠ORP are right angles. Now, in triangles OQP and ORP, OQ = OR (both are radii of the same circle), OP is common, and both triangles are right-angled. By the RHS congruence rule, ΔOQP ≅ ΔORP. Therefore, PQ = PR by CPCT (Corresponding Parts of Congruent Triangles). This proves that the lengths of the two tangents from an external point to a circle are equal.

Solution

A line can have three different positions with respect to a circle: non-intersecting, secant, and tangent. A non-intersecting line does not touch or intersect the circle at any point. A secant line intersects the circle at two distinct points, forming a chord. A tangent line touches the circle at exactly one point, known as the point of contact. These positions are determined by the distance of the line from the center of the circle relative to the radius. For example, if the distance is greater than the radius, the line is non-intersecting; if equal, it's tangent; and if less, it's a secant. Understanding these positions is fundamental in solving problems related to circles and their properties.

Solution

From a point outside the circle, exactly two tangents can be drawn to the circle. This is because the external point lies outside the circle, and the tangents must touch the circle at exactly one point each. The two tangents will be symmetric with respect to the line joining the external point and the center of the circle. The lengths of these two tangents are equal, as proven by the theorem that states the lengths of tangents drawn from an external point to a circle are equal. This property is useful in various geometric constructions and proofs.

Solution

To demonstrate the existence of a tangent at a point on a circle, perform the following activity: Take a circular wire and attach a straight wire AB at a point P of the circular wire so that it can rotate about P. Place the system on a table and gently rotate AB about P. Observe that in most positions, AB intersects the circle at P and another point. However, in one specific position, AB intersects the circle only at P. This position is the tangent to the circle at P. This activity shows that there is exactly one tangent at any point on a circle, and it is perpendicular to the radius at that point.

Solution

To prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact, consider a circle with center O and a tangent XY at point P. Take any other point Q on XY and join OQ. Since Q is outside the circle, OQ > OP (the radius). This is true for every point Q on XY except P, where OQ = OP. Therefore, OP is the shortest distance from O to XY, meaning OP is perpendicular to XY. This proves that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Solution

To find the length of a tangent drawn from a point 10 cm away from the center of a circle with radius 6 cm, use the Pythagorean theorem. Let the point be P, the center be O, and the point of contact be Q. The tangent PQ is perpendicular to the radius OQ, forming a right triangle OPQ. Here, OP = 10 cm, OQ = 6 cm. By the Pythagorean theorem, PQ² = OP² - OQ² = 10² - 6² = 100 - 36 = 64. Therefore, PQ = √64 = 8 cm. The length of the tangent is 8 cm.

Solution

A line that is perpendicular to the radius at its outer end is a tangent to the circle because it touches the circle at exactly one point. By definition, a tangent is a line that intersects the circle at only one point and is perpendicular to the radius at that point. If a line is perpendicular to the radius at a point on the circle, then any other point on the line will be at a greater distance from the center than the radius, meaning it cannot intersect the circle at any other point. Therefore, the line is a tangent to the circle at that point.

Solution

To prove this, let P be an external point, and let PA and PB be the two tangents from P to the circle with center O, touching the circle at A and B respectively. Join OA, OB, and OP. Since PA and PB are tangents, ∠OAP = ∠OBP = 90°. In quadrilateral OAPB, the sum of the angles is 360°. Therefore, ∠AOB + ∠APB + ∠OAP + ∠OBP = 360°. Substituting the known angles, ∠AOB + ∠APB + 90° + 90° = 360°, which simplifies to ∠AOB + ∠APB = 180°. This shows that the angle between the two tangents (∠APB) is supplementary to the angle subtended by AB at the center (∠AOB).

Solution

To prove that AB + CD = AD + BC for a quadrilateral ABCD circumscribing a circle, note that the points of contact divide the sides into segments that are equal in length. Let the circle touch AB at P, BC at Q, CD at R, and DA at S. By the property of tangents from an external point, AP = AS, BP = BQ, CQ = CR, and DR = DS. Therefore, AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC. This proves that the sum of one pair of opposite sides is equal to the sum of the other pair.

Solution

Consider a circle with center O and a tangent XY at point P on the circle. To prove OP is perpendicular to XY, assume a point Q on XY other than P and join OQ. Since Q is outside the circle, OQ > OP. This holds for every point on XY except P, making OP the shortest distance from O to XY, hence OP is perpendicular to XY.

Solution

Since TP = TQ, triangle TPQ is isosceles. Let angle PTQ = θ. Then, angles TPQ and TQP are each (180° - θ)/2. Angle OPT is 90° as TP is a tangent. Thus, angle OPQ = angle OPT - angle TPQ = 90° - (90° - θ/2) = θ/2. Therefore, angle PTQ = 2 * angle OPQ.

Solution

Join OT to intersect PQ at R, making PR = RQ = 4 cm. Using Pythagoras' theorem in triangle OPR, OR = √(OP² - PR²) = 3 cm. Triangles TRP and PRO are similar by AA similarity, leading to TP/PO = RP/RO. Substituting known values gives TP = 20/3 cm.

Solution

Given a circle with center O and external point P, tangents PQ and PR meet the circle at Q and R. Triangles OQP and ORP are right-angled at Q and R, respectively. They share OP and have OQ = OR (radii). By RHS congruency, the triangles are congruent, hence PQ = PR.

Solution

Let the two circles have center O. AB is a chord of the larger circle tangent to the smaller circle at P. By the tangent property, OP is perpendicular to AB. In the larger circle, the perpendicular from the center bisects the chord, hence AP = BP.

Solution

Let the circle touch AB, BC, CD, and DA at P, Q, R, and S, respectively. Since lengths of tangents from a point to a circle are equal, AP = AS, BP = BQ, CR = CQ, and DR = DS. Adding these, AB + CD = (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ) = AD + BC.

Solution

Let tangents PA and PB from external point P touch the circle at A and B. In quadrilateral OAPB, angles OAP and OBP are 90° each. Thus, angle AOB + angle APB = 180°, making them supplementary.

Solution

In a parallelogram circumscribed around a circle, opposite sides are equal and the sum of one pair of adjacent sides equals the sum of the other pair. Since all sides are equal (as tangents from a point to a circle are equal), the parallelogram must be a rhombus.

Solution

Let XY and X'Y' be parallel tangents touching the circle at P and Q, and AB be another tangent touching at C, intersecting XY at A and X'Y' at B. Triangles OAP and OAC are congruent, as are triangles OBQ and OBC, leading to angle AOB = angle AOC + angle BOC = 90°.

Solution

Let the circle touch AB at E and AC at F. Given BD = 8 cm and DC = 6 cm, so BC = 14 cm. Let AE = AF = x cm. Then, AB = x + 8 cm and AC = x + 6 cm. Using the formula for the area of the triangle in terms of its inradius, we can set up an equation to solve for x, leading to AB and AC.

Solution

The proof involves using the properties of right-angled triangles formed by the radius and tangent. For the real-life problem, consider the distance from a point outside the park to the two points of tangency, ensuring equal lengths.

Solution

The angle subtended at the minor arc is 60 degrees, and at the major arc is 120 degrees, using the properties of equilateral triangles and cyclic quadrilaterals.

Solution

The proof involves showing that the line joining the centers is perpendicular to the common chord and bisects it, using the property that the perpendicular from the center to a chord bisects the chord.

Solution

Using the Pythagorean theorem in the right-angled triangles formed by the radius and half the chord lengths, BC is found to be 9.6 cm.

Solution

The proof involves using the properties of tangents and the fact that the sum of angles in a quadrilateral is 360 degrees.

Solution

Using the property that the lengths of tangents from a point to a circle are equal, AB and AC are found to be 12 cm and 10 cm respectively.

Solution

The proof involves showing that all sides of the parallelogram are equal, using the property that the lengths of tangents from a point to a circle are equal.

Solution

The length of the chord is 8 cm, found by using the Pythagorean theorem in the right-angled triangle formed by the radius of the smaller circle, half the chord, and the radius of the larger circle.

Solution

The proof involves using the property that the sum of angles around a point is 360 degrees and that the angles subtended by the sides of the quadrilateral at the center are equal in pairs.

Solution

The lengths of the tangents are 12 cm each, and the angle between them is approximately 67.38 degrees, found using the Pythagorean theorem and trigonometric ratios.