Home
Class 12
MATHS
Let two parallel lines L1 and L2 with po...

Let two parallel lines `L_1` and `L_2` with positive slope are tangent to the circle `C_1` : `x^2 + y^2 -2x 16y + 64 = 0`. If `L_1` is also tangent to the circle `C_2` : `x^2 + y^2 - 2x + 2y -2 = 0` and the equation of `L_2` is `asqrta x - by + c - a sqrta = 0` where a,b,c in N. then find the value of `(a+b+c)/7`

Text Solution

Verified by Experts

The correct Answer is:
`2sqrt(2)x-y+11-2 sqrt(2)=0`
`S_(1)-=x^(2)+y^(2)-2x-16y+64=0`
`S_(2) -=x^(2)+y^(2)-2x+2y-2=0`
`C_(1)-=(1,8),r_(1)=1`
`C_(2)-= (1,-1),r_(2)=2`

Line `L_(1)` is transverse common tangent of `S_(1)` and `S_(2)`.
So, P divides `C_(1)C_(2)` internally in the ratio of `1:2`.
`implies P -= ((1+2)/(1+2),(-1+16)/(1+2))-=(1,5)`
Thereforem equation of `L_(1)` with positive slope is
`(y-5)=m(x-1)`
or `mx-y-m+5=0`
This is also tangnet to `S_(2)` .
So, we have
`|(m+1-m+5)/(sqrt(m^(2)+1))|=2`
`:. m=2sqrt(2)`
So, equation of `L_(1)` is `2sqrt(2)x-y-2sqrt(2)+5=0`
Thus, equation of `L_(2)` is `2sqrt(2)x -y+k=0`.
From the figure, distance between parallel lines is 2.
`:. |(k+2sqrt(2)-5)/(3)|=2`
`:. k= +- 6+5-2sqrt(2)`
Therefore, equation of `L_(2)` with `+` ve y-intercept is
`sqrt(2)x-y+11-2sqrt(2)=0`
Promotional Banner

Similar Questions

Explore conceptually related problems

Let two parallel lines L_(1) and L_(2) with positive slope are tangent to the circle C_(1): is also x^(2)+y^(2)-2x16y+64=0. If L_(1) is also tangent to the circle C_(2):x^(2)+y^(2)-2x+2y-2=0 and the equation of L_(2) is a sqrt(a)x-by+c-a sqrt(a)=0 where a,b,in N.then find the value of (a+b+c)/(7)

The equation of tangent to the circle x^(2)+y^(2)-2x+8y-23=0 having slope 3

If y=c is a tangent to the circle x^(2)+y^(2)-2x-2y-2=0, then the value of c can be

If y=c is a tangent to the circle x^(2)+y^(2)–2x+2y–2 =0 at (1, 1), then the value of c is

If y=3x+c is a tangent to the circle x^(2)+y^(2)-2x-4y-5=0, then c is equal to :

The tangents to the circle x^(2)+y^(2)-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

If L_1 and L_2, are the length of the tangent from (0, 5) to the circles x^2 + y^2 + 2x-4=0 and x^2 + y^2-y + 1 = 0 then

If a line, y = mx + c is a tangent to the circle, (x-1)^2 + y^2 =1 and it is perpendicular to a line L_1 , where L_1 is the tangent to the circle x^2 + y^2 = 8 at the point (2, 2), then :