Question

Two circles each of radius ‘5’have common tangent at (1,1) whose equation is 3x+4y-7=0, when their centres are​

Answer 1

Center is (4,5)

Step-by-step explanation:

equation of tangent is 3x + 4y - 7 = 0  ------------------(I)

radius = 5

let center of circle be (h, k )  

distance of center from tangent = radius of circle  

so  

$|\3h + 4h - 7 | = 25 \\|\sqrt(3^2 + 4^2)  | = 32$

⇒  $3\times(h)+4\times(k) - 7 = 25$

⇒ $3\times(h)+4\times(k)  = 32$

⇒ $h = \frac {(32-4k)}{3}$------------------------------------(II)  

 equation of circle whose centre is ( h, k) & radius 5 is

⇒ $(x - h)^2 + (y - k)^2 = 25$

this circle passing through the point ( 1,2 )  

⇒ $h^2 - 2h + 1 + k^2 + - 2k + 1 = 25$

⇒ $h^2 - 2h + k^2  - 2k\ - 23$ --------------(III)  

from(II) & (III)  

⇒ $(\frac{32 - 4k}{3})^2 - 2 \times {\frac{32 - 4k}{3}} + k^2 - 2k  - 23 =0$

⇒ $25^2 - 250k +625 = 0$

⇒ $k^2 - 10k + 25 = 0$

⇒ $k^2 - 5k - 5k + 25 = 0$

⇒ $(k - 5)(k - 5) = 0$

k = 5  

$So, h = \frac {32 - 5\times4}{3}$

⇒ $h = \frac{12}{3}$

h = 4

so CENTRE IS ( 4,5 )