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?

2 Answers
Sep 6, 2016

There are 22 circles satisfying the given conds. :

(1) : x^2+y^2+6x+6y+9+0(1):x2+y2+6x+6y+9+0,

and,

(2) : 49x^2+49y^2-126x+126y+81=0(2):49x2+49y2126x+126y+81=0.

Explanation:

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

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

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

rArr |k|=r rArr k=+-r|k|=rk=±r

Similarly, h=+-rh=±r

Thus, for C(h,k)C(h,k), we have to consider following 44 Cases :

(1) C(r,r), (2) C(-r,r), (3) C(-r,-r), and (4) C(r,-r)(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.!

Sep 6, 2016

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.
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