The length of the radii of two circles are 8 cm and 3 cm , the distance between the centres is 13 cm . What is the length of the direct common tangent of the two circles?

Given: B Q = 8 cm , A P = 3 cm and A B = 13 cm . The radius of a circle is perpendicular to the tangent line through its endpoint on the circle's circumference. ∴ ∠ B Q P = 90 ° Construct a line segment A C parallel to P Q . Thus, ∠ A C B = ∠ B Q P = 90 ° and B C = B Q - A P = 8 cm - 3 cm = 5 cm Therefore, △ A C B is a right-angled triangle. Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides. ∴ A B 2 = A C 2 + B C 2 ⇒ 13 2 = A C 2 + 5 2 ⇒ A C 2 = 169 - 25 ⇒ A C = 144 ⇒ A C = 12 Thus, P Q = A C = 12 cm Hence, the length of the direct common tangent of the two circles is 12 cm .

In the given figure A C is a tangent to the circle with centre O . If ∠ A D B = 55 ° , find x a n d y . Give reasons for your answers.

Given, We know that angle between the radius and the tangent at the point of contact is the right angle. ∠ A = 90 ° Also, in ∆ O B E , O B = O E = r a d i u s r In Δ A B D , ∠ A + ∠ B + ∠ A D B = 180 ° 90 ° + ∠ B + 55 ° = 180 ° ∠ B = 180 ° − 90 ° − 55 ° = 35 ° ∠ B = ∠ O E B = 35 ° a n d ∠ O E B = ∠ D E C = 35 ° (vertically opposite angles) ∠ E D C + ∠ A D E = 180 ° ∠ E D C + 55 ° = 180 ° ∠ E D C = 180 ° − 55 ° = 125 ° I n Δ E D C , ∠ D E C + ∠ E D C + x = 180 ° 35 ° + 125 ° + x = 180 ° x = 180 ° − 35 ° − 125 ° = 20 ° I n Δ A O C , ∠ A + ∠ y + ∠ x = 180 0 90 ° + y + 20 ° = 180 ° y = 180 ° − 90 ° − 20 ° = 70 ° Hence, the value of x = 20 ° a n d y = 70 ° .

In the figure given below O is the centre of the circle. If the Q R = O P and angle ∠ O R P = 20 ° . Find the value of x giving reasons.

Given, O P = Q R and O P = O T = O Q = Q R So in ∆ R Q P R Q = Q O So ∠ Q R O = ∠ Q O R = 20 ° (the same side having same opposite angle) In ∆ R Q P ∠ R Q P + ∠ Q P R + ∠ P R Q = 180 ° ∠ R Q P = 180 - 20 - 20 = 140 ° Now ∠ R Q P + O Q P = 180 ° (linear pair) ∠ O Q P = 180 - 140 = 40 ° Now in ∆ P O Q O Q = P O ( Both are radius) ∠ Q P O = ∠ O Q P = 40 ° So ∠ P O Q = 180 - ∠ O P Q - ∠ O Q P ∠ P O Q = 180 - 40 - 40 = 100 ° Now ∠ X + ∠ P O Q + ∠ R O Q = 180 ° X = 180 - 100 - 20 X = 60 ° Hence, this is the required answer.