The equation of the circle passing through (2 | vedclass

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Question

(A) Let the points be $A(2,0)$ and $B(0,4)$. The circle with minimum radius passing through two points has the line segment joining these points as it

Vedclass · Accepted Answer

$x^2 + y^2 - 2x - 4y = 0$

Vedclass · Answer

(A) Let the points be $A(2,0)$ and $B(0,4)$. The circle with minimum radius passing through two points has the line segment joining these points as its diameter. The diameter of the circle is the distance $AB = \sqrt{(0-2)^2 + (4-0)^2} = \sqrt{4 + 16} = \sqrt{20}$. The radius $r = \frac{\sqrt{20}}{2} = \sqrt{5}$. The center of the circle is the midpoint of $AB$,which is $(\frac{2+0}{2}, \frac{0+4}{2}) = (1, 2)$. The equation of the circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$. Substituting the values,we get $(x-1)^2 + (y-2)^2 = 5$. Expanding this,we get $x^2 - 2x + 1 + y^2 - 4y + 4 = 5$. Simplifying,$x^2 + y^2 - 2x - 4y = 0$.

The equation of the circle passing through $(2,0)$ and $(0,4)$ and having the minimum radius is:

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