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 #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.!

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#.
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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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