Question

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?

Answer 1

`(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]}

Answer 2

`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!