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Question

The equation of the circle passing through the point (1, –2) and having its centre on the line 2x – y – 14 = 0 and touching the line 4x + 3y – 23 = 0

A
+ + 8x + 12y + 27 = 0
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B
+ - 12y + 27 = 0
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C
+ - 8x + 12y + 27 = 0
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D
+ - 8x + 12y + 27 = 0
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Solution

The correct option is D + - 8x + 12y + 27 = 0
Let P(1, -2) and the centre be C(a, b)
Centre lies on 2x - y - 14 = 0 2a - b - 14 = 0 b = 2a - 14 (1)
Circle touches the lines 4x + 3y - 23 = 0. Lets say that this point of touching is Q.
We know that,
CQ = CP = Radius of the circle.
(a1)2+(b+2)2 = |4a + 3b 23|42 + 32
(a1)2+(2a14+2)2 = |4a+3(2a14)23|5 [From (1)]
5a22a+1+4a248a+144 = |10a65| 55a250a+145 = 5|2a13|
5a2 - 50a + 145 = (2a - 13)2 = 4a2 - 52a + 169
a2 + 2a - 24 = 0 (a + 6)(a - 4) a = 4
Case 1 : a = 4 b = 2(4) - 14 = - 6. Centre C = (4, - 6)
Radius = CP = (41)2+(6+2)2 = 9+16 = 5
The circle equation is (x4)2+(y+6)2=52 x2 + y2 - 8x + 12y + 27 = 0
Case 2 : a = - 6 b = 2(- 6) - 14 = 26 Centre C = (-6, - 26)
Radius = CP = (61)2+(26+2)2 = 49+576 = 625 = 25
The circle equation is (x+6)2+(y+26)2=252 x2 + y2 + 12x + 52y + 87 = 0
The circles are x2 + y2 - 8x + 12y + 27 = 0, x2 + y2 + 12x + 52y + 87 = 0

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