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

2 Answers
Jun 14, 2016

Answer:

#(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

Answer:

#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!