Here we will solve different types of Problems on relation between tangent and secant.
1. XP is a secant and PT is a tangent to a circle. If PT = 15 cm and XY = 8YP, find XP.
Solution:
XP = XY + YP = 8YP + YP = 9YP.
Let YP = x. Then XP = 9x.
Now, XP × YP = PT^{2}, as the product of the segments of a secant is equal to the square of the tangent.
Therefore, 9x ∙ x = 15^{2} cm^{2}
⟹ 9x^{2} = 15^{2} cm^{2}
⟹ 9x^{2} = 225 cm^{2}
⟹ x^{2} = \(\frac{225}{9}\) cm^{2}
⟹ x^{2} = 25 cm^{2}
⟹ x = 5 cm.
Therefore, XP = 9x = 9 ∙ 5 cm = 45 cm.
2. XYZ is an isosceles triangle in which XY = XZ. If N is the mid point of XZ, prove that XY = 4 XM.
Solution:
Let XY = XZ = 2x.
Then XN = \(\frac{1}{2}\)XZ = x.
XY is a secant and XN is a tangent.
Therefore, XM × XY = XN^{2} (Product of segments of secant = square of tangent).
Therefore, XM × 2x = x^{2}
⟹ XM = \(\frac{x}{2}\).
Therefore, XY = 2x = 4 ∙ \(\frac{x}{2}\) = 4XM
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