Question

Points A(-3,2) and B(1,4) are end points of the diameter of the circle. The
equation of the circle is:​

Answer 1

Given :

• Points A(-3,2) and B(1,4) are the end points of the diameter of the circle.

To Find :

• Equation of the circle

General Equation Of Circle :

$\boxed{(x-a)^2+(y-b)^2=r^2}$

Formula Applied :

$\displaystyle{\text{Mid Point Formula = }\[\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right )\]= (a,b)}$

${\text{distance Formula}= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Solution :

• To find the equation of the circle, first we need to find the centre of the circle(a,b). Since we have end points of the circle we can find the centre point of the circle using mid-point formula.

$\displaystyle\boxed{\text{Mid Point Formula = }\[\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\]= (a,b)}$

$\displaystyle{\implies\[\left(\frac{-3+2}{2},\frac{1+4}{2}\right)\]}$

$\displaystyle{\implies\[\left(\frac{-1}{2},\frac{5}{2}\right)\]}$

• Now let's find the radius of the circle. Since we have end points of the circle, we can find diametre of the circle using distance formula, and after dividing the diametre of the circle by 2 we can obtain the radius of the circle.

$\boxed{\text{distance Formula}= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

$\sf{\implies \sqrt{(2-(-3))^2+((4-1))^2}}$

$\sf{\implies \sqrt{(2+3)^2+(3)^2}}$

$\sf{\implies \sqrt{25+9}}$

$\sf{diametre=\sqrt{34}}$

$\displaystyle\sf{ radius=\frac{diametre}{2}}$

$\displaystyle\boxed{\sf{radius=\frac{\sqrt{34}}{2}}}$

• Now substitute the value of centres of the circle and radius of the circle in the general formula of circle.

$\boxed{(x-a)^2+(y-b)^2=r^2}$

$\implies\displaystyle{\[\left(x-\[\left(\frac{-1}{2}\right\])\right)\]^2+\[\left(y-\[\left(\frac{5}{2}\right)\]\right\])^2=r^2}$

$\implies \displaystyle{\[\left(x+\frac{1}{2}\right)\]^2+\[\left(y-\frac{5}{2}\right)\]^2=\[\left(\frac{34}{4}\right)\]}$

${\implies\displaystyle {\[\left(x^2+2\times x \times \frac{1}{2}\times \frac{1}{2}+\frac{1}{4}\right)\]\[\left(y^2-2\times \frac{5}{2}\times y\times \frac{25}{4}+\frac{1}{4}\right)\]=\[\left(\frac{34}{4}\right)\]}}$

${\implies\displaystyle {\[\left(x^2+ x + \frac{1}{4}\times \right)\]\[\left(y^2-5y+\frac{25}{4}\right)\]=\frac{34}{4}}}$

$\implies \displaystyle{x^2+y^2+x-5y= \frac{34}{4}-\frac{25}{4}-\frac{1}{4}}$

$\implies \displaystyle{x^2+y^2+x-5y= \frac{34-25-1}{4}}$

$\implies \displaystyle{x^2+y^2+x-5y= \frac{8}{4}}$

Required Answer

The equation of the given circle is$\displaystyle{x^2+y^2+x-5y-2=0}$

Answer 2

Answer:

The equation of the circle is x² + y² + 2x - 6y+5 = 0

Step-by-step explanation:

Given,

The endpoints of the diameter of the circle are A(-3,2) and B(1,4)

To find,

The equation of the circle

Recall the formulas

The general equation of a circle with center (h,k) and radius 'r' is given by

$(x-h)^2 + (y-k)^2 = r^2$

The coordinates of the midpoint of the line joins two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$

The distance between the  two points $(x_1,y_1)$  and  $(x_2,y_2)$  is given by

$\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

Solution:

The center of the circle is the midpoint of the endpoints of the diameter.

The endpoints of the diameter are A(-3,2) and B(1,4)

$(x_1,y_1)$ = (-3,2) and $(x_2,y_2)$ = (1,4)

The coordinates of the midpoint = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2} )$ = $(\frac{-3+1}{2},\frac{2+4}{2} )$ = (-1,3)

Hence, the center of the circle = (h,k) = (-1,3)

The radius of the circle is half the distance between the endpoints of the diameter of the circle.

Diameter = The distance between (-3,2) and (1,4)

$(x_1,y_1)$ = (-3,2) and $(x_2,y_2)$ = (1,4)

The distance between the points  $(x_1,y_1)$ = (-3,2) and $(x_2,y_2)$ = (1,4)

$\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$ = $\sqrt{(1+3)^2 +(4-2)^2}$

= $\sqrt{(4)^2 +(2)^2}$

=$\sqrt{16 +4}$

=$\sqrt{20}$

=2$\sqrt{5}$

Diameter = 2$\sqrt{5}$

The radius of the circle = $\sqrt{5}$

The general equation of a circle with center (h,k) = (-1,3)  and radius r =  $\sqrt{5}$ is given by

$(x-h)^2 + (y-k)^2 = r^2$

(x+1)² + (y-3)² = ($\sqrt{5}$)²

x² +2x+1 + y² -6y+9 = 5

x² + y² + 2x - 6y+5 = 0

∴ The equation of the circle is x² + y² + 2x - 6y+5 = 0

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