Question

Question 14 Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Class X1 - Maths -Conic Sections Page 241

Answer 1

in this way u can solve

Answer 2

Concept :- If (a, b) is the point on the circle and (h,k) is the centre of circle, then radius (r) = √{(a-h)² + (b-k)²}. Then use equation of circle is ( x - h)² + (y - k)² = r²

Here, centre = (2, 2) and point on the circle is (4, 5)
Then, radius (r) = √{(4-2)² +(5-2)²}
r = √13

Now,
equation of circle is
(x - h)² + (y - k)² = r²
(x - 2)² + (y - 2)² = √13²
x² + 4 - 4x + y² + 4 - 4y = 13
x² + y² - 4x - 4y -5 = 0

Hence, equation of circle is x² + y² - 4x - 4y -5 = 0