Question

Find the equation of a circle whose centre is (2,-1) and which passes through the point (3,6).​

Answer 1

Step-by-step explanation:

distance between centre and point gives radius,

radius=√50=5√2

if centre (2,-1)and radius=5√2 are given,(x-h)^2+(y-k)^2=r^2

equation is (x-2)^2+(y+1)^2=50

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Answer 2

Dear Student,

● Answer -

(x-2)² + (y+1)² = 50

● Explanation -

# Given -

Centre = (2, -1) and Point = (3, 6)

h = 2

k = -1

x1 = 3

y1 = 6

# Solution -

Radius of circle is given by -

r = √[(h-x1)² + (k-y1)²]

r = √[(2-3)² + (-1-6)²]

r = √(-1² + -7²)

r = √(1 + 49)

r = √50

Now, equation of circle is -

(x-h)² + (y-k)² = r²

(x-2)² + [y-(-1)]² = (√50)²

(x-2)² + (y+1)² = 50

Hence, equation of circle is (x-2)²+(y+1)² = 50.

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