Question

find the equation of circle with (4,2) and (1,5) as the two ends of the diameter and also find its centre and radius​

Answer 1

Concept

All points in a plane that are at a specific distance from a specific point, the center, form a circle. In other words, it is the curve that a moving point in a plane draws to keep its distance from a specific point constant.

Given

Circle with $(4,2)$ and $(1,5)$ as the two ends of the diameter.

To Find

We have to find the center, radius and equation of the circle.

Solution

According to the problem,

Let center of the circle be C(h,k)

Therefore, $h=\frac{4+1}{2} =\frac{5}{2}$

$k=\frac{2+5}{2} =\frac{7}{2}$

Hence center of circle is C($5/2,7/2$).

Radius of circle is $=\sqrt{(4-5/2)^{2}+(2-7/2)^{2} } =\sqrt{(3/2)^{2} +(-3/2)^{2} } =\sqrt{9/4+9/4} =3/\sqrt{2}$

Equation of circle is $(x-h)^{2} +(y-k)^{2} =r^{2}$

⇒$(x-5/2)^{2} +(y-7/2)^{2} =(3/\sqrt{2}) ^{2}$

⇒$4x^{2} +4y^{2} -20x-28y+560=0$

Hence, radius of circle is $3/\sqrt{2} units$ center of circle is C($5/2,7/2$) and equation of circle is $4x^{2} +4y^{2} -20x-28y+560=0$.

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