How do you write an equation for a circle given center in the second quadrant, tangent to y=-1, y=9 and the y-axis?

2 Answers
Sep 5, 2017

Answer:

#(x+5)^2+(y-4)^2=25#

Explanation:

If the circle is tangent to #y=-1# and #y=9#
then it has a diameter of #abs(9-(-1))=10#
which implies a radius of #5#

If the bottom of the circle is at #y=-1# and the circle has a radius of #5#,
the y-coordinate of the center of the circle must be #-1+5=4#

If the circle has a radius of #5# and it is tangent to the Y-axis (i.e. #x=0#) and it is in the second quadrant,
then the x-coordinate of the center of the circle is #0-5=-5#

So the center of the circle has coordinates #(x_c,y_c)=(-5,4)#
and its radius is #5#

The standard equation for a circle is
#color(white)("XXX")((x-x_c)^2+(y-y_c)^2=r^2#
for a circle with center #(x_c,y_c)# and radius #r#

Therefore the equation of the desired circle is
#color(white)("XXX")(x+5)^2+(y-4)^2=25#

graph{(x+5)^2+(y-4)^2=25 [-13.25, 9.25, -1.705, 9.545]}

Sep 5, 2017

Answer:

#x^2+y^2+10x-8y+16=0, or, (x+5)^2+(y-4)^2=25, #

Explanation:

We know that, the General Eqn. of a Circle is, given by,

# S : x^2+y^2+2gx+2fy+c=0," where, "g^2+f^2-c > 0.#

#"The Centre "C" and radius "r" of "S" are, resp., "C(-g,-f)#

# and sqrt(g^2+f^2-c).#

Since, #Y-# Axis, i.e., #x=0# is tgt. to #S,# the #bot-#dist. from

#C# to it equals the radius #r.#

#:. |-g|=r=sqrt(g^2+f^2-c) rArr f^2-c=0..................(1).#

Similarly, #y+1=0 and y-9=0" are tgts. to "S.#

# :. |-f+1|=r, and, |-f-9|=r.#

# rArr |-f+1|=|-f-9|.#

# rArr -f+1=-f-9," which is impossible, or, "-f+1=f+9.#

#:. -2f=8, i.e., f=-4......................(2).#

Then, by #(1), c=16...................(3).#

Also, #|-f+1|=r=sqrt(g^2+f^2-c), (2), and, (1)," give"#

#|4+1|=sqrt(g^2+0) rArr g=+-5.#

But, #g=-5, f=-4 rArr" the Centre "C(-g,-f)=C(5,4)#

lies in the First Quadrant, contrary to the Hypo.

#:. g=5, f=-4, and, c=16,# give the desired eqn. as,

#x^2+y^2+10x-8y+16=0, or, (x+5)^2+(y-4)^2=25, #

as Respected Alan P. Sir has already derived!