Question

Find the equation of the circle whose diameter is the line joining the points (-4, 3) and
(12,-1). Find also the intercept made by it on the y-axis.
The equation of the required circle is
(x + 4)(x-12)+(-3) (y + 1) = 0.
Can someone explain further about the next steps
​

Answer 1

Answer:

1 + 2√13, 1 - 2√13

Step-by-step explanation:

Points are (-4,3) and (12,-1).

Centre = (-4 + 12/2, 3 - 1/2)

           = (4,1)

Radius r = √4 - (-4)^2 + (1 - 3)^2

             = √68

Equation of circle is :

(x - p)^2 + (y - r)^2 = (√68)^2

=> (x - 4)^2 + (y - 1)^2 = √68

=> x² - 8x + 16 + y² - 2y + 1 = 68

=> x² + y² - 8x - 2y - 51 = 0    ----- (1)

We know that x = 0 on y - axis. Substituting x = 0 in (1) we get

=> y² - 2y - 51 = 0

=> y = (2 ± √208)/2

=> y = 1 ± 2√13

Hence, y-intercepts are 1 ± 2√13

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