How do you find the equation of the circle given diameter at (4, -3) and (-2, 5)?

2 Answers
Jun 14, 2016

(x-1)^2+(y-1)^2=25

Explanation:

If the diameter is color(brown)(""(4,-3)) to color(orange)(""(-2,5))
then center is at
color(white)("XXX")((color(brown)(4)+(color(orange)(-2)))/2,(color(brown)(-3)+color(orange)(5))/2)=(color(red)(1),color(blue)(1))
and
the radius is
color(white)("XXX")sqrt((color(brown)(4)-color(red)(1))^2+(color(brown)(-3)-color(blue)(1))^2)=sqrt(9+16) =color(green)(5)

The general equation of a circle
with center (color(red)(a),color(blue)(b)) and radius color(green)(r) is
color(white)("XXX")(x-color(red)(a))^2+(y-color(blue)(b))^2=color(green)(r)^2

In this case we have
color(white)("XXX")(x-color(red)(1))^2+(y-color(blue)(1))^2=color(green)(5)^2

Jun 14, 2016

x^2+y^2-2x-2y-23=0,

Explanation:

Formula for the eqn. of a circle having (x_1,y_1), (x_2,y_2) as diametrically opposite points (pts.) : (x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.

Accordingly, the reqd. eqn. is (x-4)(x+2)+(y+3)(y-5)=0,
i.e., x^2-4x+2x-8+y^2+3y-5y-15=0,
or, x^2+y^2-2x-2y-23=0,

Aliter :-

Extremities of a diameter are (4,-3), (-2,5)

Centre, being midpoint of diameter, we get the centre ((4-2)/2,(-3+5)/2)=(1,1).

Radius = Distance between the Centre & any end point of given diameter = sqrt {(4-1)^2+(-3-1)^2}=5.

Hence the eqn. : (x-1)^2+(y-1)^2=5^2, or, x^2+y^2-2x-2y-23=0, as before!