Question

A circle that has (-2,4) and (4,2) as endpoints of a diameter.

Use the given information to write the standard equation of the circle.

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Answer 1

Answer:

$\implies (x-1)^2+(y-3)^2=10$

Step-by-step explanation:

End points of the diameter of the circle are:

$(-2,\:4) \implies x_1= -2, \: y_1= 4$

$(4,\:2) \implies x_2= 4, \: y_2= 2$

Equation of circle in diameter form is given as:

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

Plugging in the values of $x_1, \: y_1,\:x_2, \: y_2$ in the above equation, we find:

$[x-(-2)](x-4)+(y-4)(y-2)=0$

$\implies (x+2)(x-4)+(y-4)(y-2)=0$

$\implies x^2-2x-8+y^2-6y+8=0$

$\implies x^2-2x+1-9+y^2-6y+9-1=0$

$\implies (x^2-2x+1)-9+(y^2-6y+9)-1=0$

$\implies (x-1)^2+(y-3)^2-10=0$

$\implies (x-1)^2+(y-3)^2=10$

This is the required equation of circle in standard form.