# How do you find the equation of a circle whose center of this circle is on the line 2x-5y=9 and it is tangent to both the x and y axis?

Sep 6, 2016

There are $2$ circles satisfying the given conds. :

$\left(1\right) : {x}^{2} + {y}^{2} + 6 x + 6 y + 9 + 0$,

and,

$\left(2\right) : 49 {x}^{2} + 49 {y}^{2} - 126 x + 126 y + 81 = 0$.

#### Explanation:

Let $C \left(h , k\right)$ be the centre, and $r > 0$, the radius.

The Circle touches X-axis [eqn. $y = 0$] . By Geom., we have, then,

$\text{The" bot-"dist. btwn. "C(h,k) and X"-axis=} r$.

$\Rightarrow | k | = r \Rightarrow k = \pm r$

Similarly, $h = \pm r$

Thus, for $C \left(h , k\right)$, we have to consider following $4$ Cases :

$\left(1\right) C \left(r , r\right) , \left(2\right) C \left(- r , r\right) , \left(3\right) C \left(- r , - r\right) , \mathmr{and} \left(4\right) C \left(r , - r\right)$

Since, $C \left(h , k\right) \text{ lies on the line : } 2 x - 5 y = 9 \therefore 2 h - 5 k = 9. \ldots \left(\star\right) .$.

Case (1) C(r,r) :=

$\left(\star\right) \Rightarrow 2 r - 5 r = 9 \Rightarrow r = - 3 \text{, not possible, as } r > 0$.

Case (2) C(-r,r) :=

$\left(\star\right) \Rightarrow - 2 r - 5 r = 9 \Rightarrow r = - \frac{9}{7} \text{, not possible, as } r > 0$.

Case (3) C(-r,-r) :=

$\left(\star\right) \Rightarrow - 2 r + 5 r = 9 \Rightarrow r = 3 > 0$

Thus, the Centre is $C \left(- 3 , - 3\right) , \mathmr{and} , r = 3 \Rightarrow$ The eqn. is

${\left(x + 3\right)}^{2} + {\left(y + 3\right)}^{2} = {3}^{2} , i . e . , {x}^{2} + {y}^{2} + 6 x + 6 y + 9 + 0$.

Case (4) : C(r,-r) :=

By $\left(\star\right) , 2 r + 5 r = 9 \Rightarrow r = \frac{9}{7} > 0 \Rightarrow C \left(\frac{9}{7} , - \frac{9}{7}\right) , r = \frac{9}{7}$.

$\Rightarrow \text{The eqn. : } {\left(x - \frac{9}{7}\right)}^{2} + {\left(y + \frac{9}{7}\right)}^{2} = {\left(\frac{9}{7}\right)}^{2}$, i.e.,

$49 {x}^{2} + 49 {y}^{2} - 126 x + 126 y + 81 = 0$.

Enjoy Maths.!

Sep 6, 2016

There are two circles in Q3 and Q4. They are given by
${\left(x - \frac{9}{7}\right)}^{2} + {\left(y + \frac{9}{7}\right)}^{2} = {\left(\frac{9}{7}\right)}^{2} \mathmr{and} {x}^{2} + {y}^{2} + 6 x + 6 y + 9 = 0$.
,

#### Explanation:

The given line makes intercepts $\frac{9}{2} \mathmr{and} - \frac{9}{5}$ on the axes. So, the

the second quadrant Q2 is out.

As the circle touches the axes, the equation has the form

${\left(x \pm a\right)}^{2} + {\left(y \pm a\right)}^{2} = {a}^{2} , a > 0$. Excluding Q2, .

the center $\left(a , a\right) \mathmr{and} \left(a , - a\right) \mathmr{and} \left(- a , - a\right)$ lies on $2 x - 5 y = 9$.

Negative a from the first is ruled out. So,

from the second and third,

a = 9/7, for the circle in Q4 and

a = 3, for the circle in Q3. .

Thus, there are two circles in Q3 and Q4. They are given by

${\left(x - \frac{9}{7}\right)}^{2} + {\left(y + \frac{9}{7}\right)}^{2} = {\left(\frac{9}{7}\right)}^{2}$ and ${x}^{2} + {y}^{2} + 6 x + 6 y + 9 = 0$.
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