Question

2 circles of radius 5 units touch each other at 1,2 of tbe equation of their common tangent is 4x+3y=10

Answer 1

hope it helps you and the others

Answer 2

Answer:

(5,5) and (−3,−1)

Step-by-step explanation:

Let (a,b) be the centre of on of the circle.

Then, centre must lie on the line ⊥ to the common tangent 4x+3y=10 and passing through the point (1,2). Thus, the equation of the radius is  

3x −4y + k = 0          ...(1)

Since it passes through (1,2), therefore

3 − 8 + k = 0 i.e., k = 5

Substituting k=5 in (1), the equation of the line joining centres is  

3x − 4y + 5 = 0       ...(2)

As center lies on (2), we have 3a−4b+5=0

⇒b = (3a + 5)/4...(3)

Since the radius of circle is 5, therefore  

(a−1)²  + (b−2)² =25   ...(4)

Substituting the value of b from (3) in (4), we get  

(a − 1)² + ( (3a + 5)/4 - 2)² = 25

⇒(a − 1)² = ±4

⇒a = 5 or a = −5   ...(5)

From (3)and (5), we have

a = 5 , b = 5

or

a = −3 , b = −1.

Thus, the centers of the circle are (5,5) and (−3,−1)