Question

find the equations of the circles which have radius √13 and touch the line 2x - 3 Y + 1 = 0 at (1, 1)​

Answer 1

Step-by-step explanation:

Given  

find the equations of the circles which have radius √13 and touch the line 2x - 3 Y + 1 = 0 at (1, 1)

• We know that (x – x1)^2 + (y – y1)^2 = 13  
• So x1y1 are centres
• So now the equation will be
• (1 – x1)^2 + (1 – y1)^2 = 13-----------------1
• Slope of 2x – 3y + 1 = 0 is 3y = 2x + 1
•                        Or y = 2/3 x + 1/3
• This is in the form of y = mx + c and so slope will be 2/3 and perpendicular to slope is – 3/2
• So y1 – 1 / x1 – 1 = - 3/2
• So (y1 – 1 ) = - 3/2 (x1 – 1)
• (1 – y1) = - 3/2 (1 – x1) squaring we get
• (1 – y1)^2 = 9/4 (1 – x1)^2
• From eqn 1 we get
• (1 – x1)^2 + 9/4 (1 – x1)^2 = 13
• (1 – x1)^2 [ 1 + 9/4] = 13
• (1 – x1)^2 = 4
• So (1 – x1) = 2, - 2
• So x1 = - 1, x1 = 3
• Now y1 – 1 = - 3/2(x – 1)
• Or y1 = -3/2 (x – 1) + 1
• Now y1 = - 3/2(-1 – 1) + 1
• Or y1 = - 3/2 (- 2) + 1
• So y1 = 4
• Also y1 = - 3/2 (3 – 1) + 1
•             = - 2
• Now we get (- 1, 4) and (3, - 2)
• Given equation of circle is (x – x1)^2 + (y – y1)^2 = 13

Also (x + 1)^2 + (y – 4)^2 = 13----------------1

      (x – 3)^2 + (y + 2)^2 = 13 ----------------2

Now these are the two equations  

Reference link will be

https://brainly.in/question/8657615

Answer 2

Answer:

Step-by-step explanation: