Question

find the equation of circle whose center (1,2) & which passes through the point (-2,3)​

Answer 1

Step-by-step explanation:

Given: Centre is (1, 2) and which passes through the point (4, 6).

Where, p = 1, q = 2

We need to find the equation of the circle.

By using the formula,

(x – p)2 + (y – q)2 = r2

(x – 1)2 + (y – 2)2 = r2

It passes through the point (4, 6)

(4 – 1)2 + (6 – 2)2 = r2

32 + 42 = r2

9 + 16 = r2

25 = r2

r = √25 = 5

So r = 5 units

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2

By substitute the values in the above equation, we get

(x – 1)2 + (y – 2)2 = 52

x2 – 2x + 1 + y2 – 4y + 4 = 25

x2 + y2 – 2x – 4y – 20 = 0.

∴ The equation of the circle is x2 + y2 – 2x – 4y – 20 = 0.