1. From point Q, the length of the tangent to a circle is 24 cm, and the distance of Q from the center is 25 cm. The radius of the circle is
(A) 7 cm.
(B) 12 cm.
(C) 15 cm.
(D) 24.5 cm.
First, draw a perpendicular from the center O of the triangle to a point P on the circle, which is touching the tangent. This line will be perpendicular to the tangent of the circle.
So, OP is perpendicular to PQ, i.e., OP ⊥ PQ
From the above figure, it is also seen that △OPQ is a right-angled triangle.
It is given that
OQ = 25 cm and PQ = 24 cm
By using Pythagoras’ theorem in △OPQ,
$OQ^{2}$ = $OP^{2} +PQ^{2}$
$(25)^{2}$ = $OP^{2}+(24)^{2}$
$OP^{2}$ = 625-576
$OP^{2}$ = 49
OP = 7 cm
So, option A, i.e., 7 cm, is the radius of the given circle.
(A) 60°
(B) 70°
(C) 80°
(D) 90°
From the question, it is clear that OP is the radius of the circle to the tangent PT, and OQ is the radius to the tangent TQ.
So, OP ⊥ PT and TQ ⊥ OQ
∴ ∠OPT = ∠OQT = 90°
Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360°.
So, ∠PTQ+∠POQ+∠OPT+∠OQT = 360°
Now, by putting the respective values, we get
∠PTQ +90°+110°+90° = 360°
∠PTQ = 70°
So, ∠PTQ is 70° which is option B.
(A) 50°
(B) 60°
(C) 70°
(D) 80°
First, draw the diagram according to the given statement.
Now, in the above diagram, OA is the radius to tangent PA, and OB is the radius to tangent PB.
So, OA is perpendicular to PA, and OB is perpendicular to PB, i.e., OA ⊥ PA and OB ⊥ PB.
So, ∠OBP = ∠OAP = 90°
Now, in the quadrilateral AOBP,
The sum of all the interior angles will be 360°.
So, ∠AOB+∠OAP+∠OBP+∠APB = 360°
Putting their values, we get
∠AOB + 260° = 360°
∠AOB = 100°
Now, consider the triangles △OPB and △OPA. Here,
AP = BP (Since the tangents from a point are always equal)
OA = OB (Which are the radii of the circle)
OP = OP (It is the common side)
Now, we can say that triangles OPB and OPA are similar using SSS congruency.
∴ △OPB ≅ △OPA
So, ∠POB = ∠POA
∠AOB = ∠POA+∠POB
2 (∠POA) = ∠AOB
By putting the respective values, we get
=>∠POA = $\dfrac{100°}{2}$ = 50°
As the angle ∠POA is 50°, option A is the correct option.
First, draw a circle and connect two points, A and B, such that AB becomes the diameter of the circle. Now, draw two tangents, PQ and RS, at points A and B, respectively.
Now, both radii, i.e. AO and OB, are perpendicular to the tangents.
So, OB is perpendicular to RS, and OA is perpendicular to PQ.
So, ∠OAP = ∠OAQ = ∠OBR = ∠OBS = 90°
From the above figure, angles OBR and OAQ are alternate interior angles.
Also, ∠OBR = ∠OAQ and ∠OBS = ∠OAP (Since they are also alternate interior angles)
So, it can be said that line PQ and line RS will be parallel to each other (Hence Proved).
Let, O is the center of the given circle.
A tangent PR has been drawn touching the circle at point P.
Draw QP ⊥ RP at point P, such that point Q lies on the circle.
∠OPR = 90° (radius ⊥ tangent)
Also, ∠QPR = 90° (Given)
∴ ∠OPR = ∠QPR
Now, the above case is possible only when centre O lies on the line QP.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the center of the circle.
Draw the diagram as shown below.
Here, AB is the tangent that is drawn on the circle from point A.
So, the radius OB will be perpendicular to AB, i.e., OB ⊥ AB
We know, OA = 5cm and AB = 4 cm
Now, In △ABO,
$OA^{2}$ =$AB^{2}+BO^{2} $(Using Pythagoras’ theorem)
$5^{2}$ = $4^{2}+BO^{2}$
$BO^{2} $= 25-16
$BO^{2}$ = 9
BO = 3
So, the radius of the given circle, i.e., BO, is 3 cm.
Draw two concentric circles with the center O. Now, draw a chord AB in the larger circle, which touches the smaller circle at a point P, as shown in the figure below.
From the above diagram, AB is tangent to the smaller circle to point P.
∴ OP ⊥ AB
Using Pythagoras’ theorem in triangle OPA,
OA$^{2}$= AP$^{2}$+OP$^{2}$
5$^{2}$ = AP$^{2}$+3$^{2}$
AP$^{2}$ = 25-9
AP = 4
Now, as OP ⊥ AB,
Since the perpendicular from the center of the circle bisects the chord, AP will be equal to PB.
So, AB = 2AP = 2×4 = 8 cm
So, the length of the chord of the larger circle is 8 cm.
The figure given is:
From the figure, we can conclude a few points, which are
(i) DR = DS
(ii) BP = BQ
(iii) AP = AS
(iv) CR = CQ
Since they are tangents on the circle from points D, B, A, and C, respectively.
Now, adding the LHS and RHS of the above equations, we get,
DR+BP+AP+CR = DS+BQ+AS+CQ
By rearranging them, we get,
(DR+CR) + (BP+AP) = (CQ+BQ) + (DS+AS)
By simplifying,
AD+BC= CD+AB
From the figure given in the textbook, join OC. Now, the diagram will be as
Now, the triangles △OPA and △OCA are similar using SSS congruency as
(i) OP = OC They are the radii of the same circle
(ii) AO = AO It is the common side
(iii) AP = AC These are the tangents from point A
So, △OPA ≅ △OCA
Similarly,
△OQB ≅ △OCB
So,
∠POA = ∠COA … (Equation i)
And, ∠QOB = ∠COB … (Equation ii)
Since the line POQ is a straight line, it can be considered as the diameter of the circle.
So, ∠POA +∠COA +∠COB +∠QOB = 180°
Now, from equations (i) and equation (ii), we get,
2∠COA+2∠COB = 180°
∠COA+∠COB = 90°
∴ ∠AOB = 90°
First, draw a circle with center O. Choose an external point P and draw two tangents, PA and PB, at point A and point B, respectively.
Now, join A and B to make AB in a way that subtends ∠AOB at the center of the circle. The diagram is as follows:
From the above diagram, it is seen that the line segments OA and PA are perpendicular.
So, ∠OAP = 90°
In a similar way, the line segments OB $bot$ PB and so, ∠OBP = 90°
Now, in the quadrilateral OAPB,
∴ ∠APB+∠OAP +∠PBO +∠BOA = 360° (since the sum of all interior angles will be 360°)
By putting the values, we get,
∠APB + 180° + ∠BOA = 360°
So, ∠APB + ∠BOA = 180° (Hence proved).