A tangent is a straight line that touches the circumference of a circle at only one point. The angle between a tangent and the radius is 90˚.
Problem 1 :
Length of the tangent is 8 cm, radius of the circle is 6 cm. Find the distance from center to the point away from the circle.
(a) 10 cm (b) 5 cm (c) √7 cm (d) 2√7 cm
Solution :
OP is perpendicular to TP.
TO^{2} = TP^{2} + OP^{2}
TO^{2} = 8^{2} + 6^{2}
TO^{2} = 64 + 36
TO^{2} = 100
TO = √100
TO = 10 cm
Problem 2 :
If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of 80°, then ∠POA is
(a) 30^{0 } (b) 60^{0 } (c) 50^{0} (d) 100^{0}
Solution :
OA is perpendicular to AP.
∠POA = ?
∠POA + ∠APO + ∠PAO = 180
∠POA + 40 + 90 = 180
∠POA + 130 = 180
∠POA = 180 - 130
∠POA = 50
Problem 3 :
If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm then length of each tangent is --------------
Solution :
OA is perpendicular to PA. To find the length of tangent PA, we use
tan θ = Opposite side/Adjacent side
tan 30 = OA/PA
1/√3 = 3/PA
PA = 3√3
Problem 4 :
If the angle between two tangents drawn from a point T to a circle of radius r and center O is 120^{0} then find the length of TP .
Solution :
OA is perpendicular to PA. To find the length of tangent PA, we use
tan θ = Opposite side/Adjacent side
tan 60 = OP/TP
√3 = r/TP
TP = r/√3
Problem 5 :
The length of a tangent from a point A at distance 5 cm from the center of the circle is 4 cm. Find the radius of the circle.
(a) 4 cm (b) 3 cm (c) 6 cm (d) 5 cm
Solution :
Length of tangent = 4 cm
Distance from the point away from the circle to center = 5 cm
The line drawn from center to tangent must be a perpendicular.
5^{2} = 4^{2} + r^{2}
25 = 16 + r^{2}
r^{2} = 25 - 16
r^{2} = 9
r = 3 cm
Problem 6 :
PQ is a chord of length 8 cm of a circle of radius 5 cm .The tangents at P and Q intersect at a point T . Find the length TP.
Solution :
Length of PQ = 8 cm, when we draw the perpendicular from O to PQ, it will be a perpendicular bisector.
The midpoint of PQ be C.
PO^{2} = OR^{2} + RP^{2}
5^{2} = OR^{2} + 4^{2}
25 - 16 = OR^{2}
OR^{2} = 9
OR = 3
In triangle TPR,
TP^{2} = TR^{2} + RP^{2}
OS = 5 cm, OS = SR + RO
5 = SR + 3
SR = 2, let TS = x and TP = y
y^{2} = (2 + x)^{2} + 4^{2}
y^{2} = 2^{2} + 2(2)x + x^{2} + 4^{2}
y^{2} = x^{2 }+ 4x + 20 -----(1)
In triangle TPO
TO^{2} = TP^{2} + OP^{2}
(5 + x)^{2} = y^{2} + 5^{2}
25 + x^{2} + 10x = y^{2} + 25
y^{2} = x^{2} + 10x -----(2)
(1) = (2)
x^{2 }+ 4x + 20 = x^{2} + 10x
6x = 20
x = 5/3
By applying the value of x in (2), we get
y^{2} = (5/3)^{2} + 10(5/3)
y^{2} = 25/9 + 50/3
y^{2} = (25+150)/9
y^{2} = 175/9
y = 4.40 cm
So, length of tangent TP is 4.40 cm.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM