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Question
Answer the following :
Find the length of the tangent segment drawn from the point (5, 3) to the circle x2 + y2 + 10x – 6y – 17 = 0
Solution 1
Given equation of circle is
x2 + y2 + 10x – 6y – 17 = 0
Comparing this equation with
x2 + y2 + 2gx + 2fy + c = 0, we get
2g = 10, 2f = –6, c = –17
∴ g = 5, f = –3, c = –17
Centre of circle = (– g, – f )
= C(– 5, 3)
Radius of circle = `sqrt("g"^2 + "f"^2 - "c")`
= `sqrt(5^2 + (-3)^2 - (-17))`
= `sqrt(25 + 9 + 17)`
= `sqrt(51)`
BC = `sqrt((-5 - 5)^2 + (3 - 3)^2`
= `sqrt(100 + 0)`
= 10
In right angled ΔABC
BC2 = AB2 + AC2 …[Pythagoras theorem]
∴ (10)2 = `"AB"^2 + (sqrt(51))^2`
∴ AB2 = 100 – 51 = 49
∴ AB = 7
∴ Length of the tangent segment from (5, 3) is 7 units.
Solution 2
Given equation of circle is
x2 + y2 + 10x – 6y – 17 = 0
Here, g = 5, f = –3, c = –17
Length of the tangent segment to the circle
x2 + y2 + 2gx + 2fy + c = 0 from the point
(x1, y1) is `sqrt(x_1^2 + y_1^2 + 2"g"x_1 + 2"f"y_1 + "c")`.
∴ Length of the tangent segment from (5, 3)
= `sqrt((5)^2 + (3)^2 + 10(5) - 6(3) - 17)`
= `sqrt(25 + 9 + 50 - 18 - 17)`
= `sqrt(49)`
= 7 units
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