Find the equation of the tangents to the circle x2 + y2-22x - 4y + 25 = 0, which are perpendicular to the straight line 5x + 12y + 9 = 0.
Answers
Answer:
12x - 5y - 8 = 0 and 12x - 5y - 242 = 0
Step-by-step explanation:
Any line perpendicular to 5x + 12y = -9 is 12x - 5y + k = 0.
This line would be a tangent to the circle if its distance from the centre of circle is equal to its radius.
(a)
2g = -22
g = -11
(b)
2f = -4
f = -2
(c)
r = √g² + f² - c
= √100
= 10
Centre of the given circle is (11,2) and radius 10.
=> [|12 * 11 - 5 * 2 + k|/√12² + 5²] = 10
=> |122 + k|/13 = 10
=> |122 + k| = 130
=> 112 + k = ±130
(i)
112 + k = 130
k = 8
(ii)
112 + k = -130
k = -242
Hence, the required tangents are,
12x - 5y - 8 = 0 and 12x - 5y - 242 = 0
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Answer:
c = 58
C= 5−252
Put value of c in (1)
5y=12x+8 and 5y=12x−252 are lines ⊥ to 5x+12y+8=0 and are tangent to given will
Step-by-step explanation:
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