# 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?

Sep 5, 2017

${\left(x + 5\right)}^{2} + {\left(y - 4\right)}^{2} = 25$

#### Explanation:

If the circle is tangent to $y = - 1$ and $y = 9$
then it has a diameter of $\left\mid 9 - \left(- 1\right) \right\mid = 10$
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 $\left({x}_{c} , {y}_{c}\right) = \left(- 5 , 4\right)$
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 $\left({x}_{c} , {y}_{c}\right)$ and radius $r$

Therefore the equation of the desired circle is
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 5\right)}^{2} + {\left(y - 4\right)}^{2} = 25$

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

Sep 5, 2017

${x}^{2} + {y}^{2} + 10 x - 8 y + 16 = 0 , \mathmr{and} , {\left(x + 5\right)}^{2} + {\left(y - 4\right)}^{2} = 25 ,$

#### Explanation:

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

$S : {x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0 , \text{ where, } {g}^{2} + {f}^{2} - c > 0.$

$\text{The Centre "C" and radius "r" of "S" are, resp., } C \left(- g , - f\right)$

$\mathmr{and} \sqrt{{g}^{2} + {f}^{2} - c} .$

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

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

$\therefore | - g | = r = \sqrt{{g}^{2} + {f}^{2} - c} \Rightarrow {f}^{2} - c = 0. \ldots \ldots \ldots \ldots \ldots . . \left(1\right) .$

Similarly, $y + 1 = 0 \mathmr{and} y - 9 = 0 \text{ are tgts. to } S .$

$\therefore | - f + 1 | = r , \mathmr{and} , | - f - 9 | = r .$

$\Rightarrow | - f + 1 | = | - f - 9 | .$

$\Rightarrow - f + 1 = - f - 9 , \text{ which is impossible, or, } - f + 1 = f + 9.$

$\therefore - 2 f = 8 , i . e . , f = - 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right) .$

Then, by $\left(1\right) , c = 16. \ldots \ldots \ldots \ldots \ldots \ldots \left(3\right) .$

Also, $| - f + 1 | = r = \sqrt{{g}^{2} + {f}^{2} - c} , \left(2\right) , \mathmr{and} , \left(1\right) , \text{ give}$

$| 4 + 1 | = \sqrt{{g}^{2} + 0} \Rightarrow g = \pm 5.$

But, $g = - 5 , f = - 4 \Rightarrow \text{ the Centre } C \left(- g , - f\right) = C \left(5 , 4\right)$

lies in the First Quadrant, contrary to the Hypo.

$\therefore g = 5 , f = - 4 , \mathmr{and} , c = 16 ,$ give the desired eqn. as,

${x}^{2} + {y}^{2} + 10 x - 8 y + 16 = 0 , \mathmr{and} , {\left(x + 5\right)}^{2} + {\left(y - 4\right)}^{2} = 25 ,$

as Respected Alan P. Sir has already derived!