how to find the radius of a circle if the two tangents 3x-4y+4=0 & 6x-8y-7=0 to a circle are given......??

L₁ = 3x - 4y + 4 = 0

L₂ = 6x - 8y -7 = 0

both of  these lines are  parallel , coz slopes are same ...

now , if two lines touch the circle & also they are parallel then surely these lines pass through

dimetrically opposite points  ... distance bw these two points is length of diameter ....

now we have to find the distance bw these two lines , let it be d then

d = [2(3x-4y)-7]/5                                 (by using distance bw two parallel lines formula)

  =  [2(-4) - 7]/5 = -3 or 3 units

radius = d/2 = 3/2 units

3x-4y+4=0
Put x = 1
3 (1) - 4y + 4 = 0
- 4y = -7
   y  = 7/4 
then, P(1, 7/4).......(1)

6x-8y-7=0
Again put x = 1
6(1) - 8y = 7
-8y = 7 - 6
y = -1/8
then, Q(1, -1/8)..........(2)

Hence the coordinates of the tangents are P(1, 7/4) & Q(1, -1/8).
From eq. (1) and (2)
find the coordinates of the centre by the mid pt. theorem
 [PO=OQ ; Radius of the circle]

x = 1+1/2 = 2/2 = 1
y = 7/4 + (-1/8) = 13/16
Hence we have now centre coordinates O (1, 13/16)

nOW by the distance formulae,
PO² = OQ²
(1-1)² + (7/4 -13/16)^{2 =} (1-1)² + (13/16 +1/8)²

(0)² + (15/16)² = (0)² + (15/16)²

(15/16)² = (15/16)²

15/16 = 15/16                                                    {PO=OQ} {RADIUS OF THE CIRCLES}

Hence, 15/16 is the radius of the circle.

.

Please Approve!

thnx