Question

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?

Answer 1

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

`(1) : x^2+y^2+6x+6y+9+0`,

and,

`(2) : 49x^2+49y^2-126x+126y+81=0`.

Explanation:

Let `C(h,k)` be the centre, and `r>0`, the radius.

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

`"The" bot-"dist. btwn. "C(h,k) and X"-axis="r`.

`rArr |k|=r rArr k=+-r`

Similarly, `h=+-r`

Thus, for `C(h,k)`, we have to consider following `4` Cases :

`(1) C(r,r), (2) C(-r,r), (3) C(-r,-r), and (4) C(r,-r)`

Since, `C(h,k) " lies on the line : "2x-5y=9 :. 2h-5k=9....(star).`.

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

`(star) rArr 2r-5r=9 rArr r=-3", not possible, as "r>0`.

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

`(star) rArr -2r-5r=9 rArr r=-9/7", not possible, as "r>0`.

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

`(star) rArr -2r+5r=9 rArr r=3>0`

Thus, the Centre is `C(-3,-3), and, r=3 rArr` The eqn. is

`(x+3)^2+(y+3)^2=3^2, i.e., x^2+y^2+6x+6y+9+0`.

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

By `(star), 2r+5r=9 rArr r=9/7>0 rArr C(9/7,-9/7), r=9/7`.

`rArr "The eqn. : "(x-9/7)^2+(y+9/7)^2=(9/7)^2`, i.e.,

`49x^2+49y^2-126x+126y+81=0`.

Enjoy Maths.!

Answer 2

There are two circles in Q3 and Q4. They are given by
`(x-9/7)^2+(y+9/7)^2=(9/7)^2 and x^2+y^2+6x+6y+9=0`.
,

Explanation:

The given line makes intercepts `9/2 and -9/5` on the axes. So, the

the second quadrant Q2 is out.

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

`(x+-a)^2+(y+-a)^2=a^2, a > 0`. Excluding Q2, .

the center `(a, a) or (a, -a) or (-a, -a)` lies on `2x-5y=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

`(x-9/7)^2+(y+9/7)^2=(9/7)^2` and `x^2+y^2+6x+6y+9=0`.
,