Mathematics

Mathematics

Problem

The equation of the circle passing through (4, 5) having the centre at (2, 2) is a. x2+y2+4x+4y−5=0x^2 + y^2 + 4x + 4y - 5 = 0 b. x2+y2−4x−4y−5=0x^2 + y^2 - 4x - 4y - 5 = 0 c. x2+y2−4x−13=0x^2 + y^2 - 4x - 13 = 0 d. x2+y2−4x−4y+5=0x^2 + y^2 - 4x - 4y + 5 = 0
Intuitive explanation

The key here is recognizing that a circle's equation is completely determined by its center and any point on it. Since you know the center (2,2)(2, 2) and a point (4,5)(4, 5) that lies on the circle, you can find the radius using the distance formula—that's the distance from center to the point. Once you have the radius rr, the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 gives you the equation immediately.

Intuitive explanation

The key here is recognizing that a circle's equation is completely determined by its center and any point on it. Since you know the center (2,2)(2, 2) and a point (4,5)(4, 5) that lies on the circle, you can find the radius using the distance formula—that's the distance from center to the point. Once you have the radius rr, the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2 gives you the equation immediately.

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