Find the equation of the circle which touches the x-axis at a distance of +5 unit from the origin and cuts off an intercept of length 24 unit from the y-axis?

3 Answers
Somebody N.
Jun 6, 2018

#color(blue)((x-5)^2+(y-601/48)^2=361201/2304)#

Explanation:

The standard equation of a circle is given by:

#(x-h)^2+(y-k)^2=r^2#

Where #bbh# and #bbk# are the #bbx# and #bby# coordinates of the centre respectively and #bbr# is the radius.

If I am reading this correctly, then " touches the x axis" means the line #y=0# is a tangent to the circle. We can therefore make the assumption that if this is the case, then the #bby# coordinate of the centre, this is #bbk# must be on the line #x=5#. This makes sense, because the radius of the circle is perpendicular to the tangent line at the point of tangency, and since the tangent line is the horizontal axis and the radius is perpendicular we know the #bby# coordinate must lie on this vertical line. This being the case:

the #x# coordinate is #5#

The #y# coordinate is #k#

centre is #(5,k)#

We know two points on the circumference, and the distance of these to the centre must be the radius:

Using the distance formula, we get:

#|d|=sqrt((5-5)^2+(k-0)^2)#

And:

#|d|=sqrt((5-0)^2+(k-24)^2)#

Equating these together:

#(5-5)^2+(k-0)^2=(5-0)^2+(k-24)^2#

#k^2-25-k^2+48k-576=0#

#48k=601#

#k=601/48#

So coordinates of centre are:

#(5,601/48)#

To find the radius we just use the distance formula with the centre and one of the points on the circumference:

i.e.

#(5,601/48) and (5,0)#

#|r|=sqrt((5-5)^2+(601/48-0)^2)=sqrt(361201/2304)#

So equation of circle is:

#(x-5)^2+(y-601/48)^2=361201/2304#

PLOT:

enter image source here

Ratnaker Mehta · Somebody N.
Jun 7, 2018

# x^2+y^2-10x+-26y+25=0#.

Explanation:

Let, # S : x^2+y^2+2gx+2fy+c=0, (sqrt(g^2+f^2-c) gt 0)# be

the reqd. eqn. of the circle.

Clearly, the centre #C# and the radius #r# of #S# are given

by, #C=C(-g,-f) and r=sqrt(g^2+f^2-c#.

Given that, #S# touches the #X"-Axis"#.

Hence, the #bot"-distance"# from #C(-g,-f)# to the #X"-Axis,"# must

equal #r#.

#:. |-f|=sqrt(g^2+f^2-c), [=r]. :. g^2-c=0............(ast^1)#.

The point of tangency , i.e., #(5,0) in S#.

#:. 5^2+0^2+2g(5)+2f(0)+c=0, or, 10g+c+25=0#.

Utilising #(ast^1)# in this, we get, #g^2+10g+25=0#.

#:. g=-5," and by "(ast^1), c=25#.

Finally, to find the constant #f#, we utilise the fact that #S# cuts an

intercept of length #24# on the #Y"-Axis"#.

To this end, suppose that,

# SnnY"-Axis"=Snn{(x,y)|x=0}={C,D}. :. CD=24#.

Then, letting #x=0" in "S#, we have, #y^2+2fy+c=0#.

If #y_1 and y_2# are the roots of this quadratic eqn. in

#y#, then clearly, #C=C(0,y_1) and D=D(0,y_2)#.

Since, #y_1+y_2=-2f and y_1*y_2=c=25#, we find,

#(y_1-y_2)^2=(y_1+y_2)^2-4y_1*y_2=(-2f)^2-100#.

But, #|y_1-y_2|=CD=24#.

#:. 4f^2-100=24^2=576 :. 4f^2=676 :. f^2=169#.

#:. f=+-13#.

Altogether, in #S," we have, "g=-5, f=+-13, c=25#.

#:. S : x^2+y^2-10x+-26y+25=0#, is the desired eqn.

Graph (Red Circles) Courtesy : Respected Somebody N.

graph{x^2+y^2-10x+26y+25=0 [-52, 52, -26, 26]} graph{x^2+y^2-10x-26y+25=0 [-52, 52, -26, 26]}

enter image source here

enter image source here

Ratnaker Mehta
Jun 9, 2018

Here is another Solution using Geometry.

# x^2+y^2-10x+-26y+25=0#.

Explanation:

Suppose that the reqd. circle #S# has centre #C# and

radius #r gt 0#.

It is given that #S# touches the #X"-Axis"# at a distance

#+5# units from the Origin #O(0,0)#.

So, the tangent (tgt.), i.e., the #X"-Axis,"# touches the circle #S# at the

point, say #A,# where, #A=A(5,0)#.

From Geometry, we know that the centre #C# must be at a

distance #r# on a line #bot# to the tgt. and passing

through the point of contact .

We conclude that, the centre #C# must be #C(5,+-r)#.

Let #BD# be the intercept cut off by #S# on the #Y"-Axis"#.

Let #M# be the mid-point of #BD.#

Then, from Geometry, #CM bot BD, &, CB=CD=r#.

Also, #CM=AO=5, BD=24, MB=1/2*BD=12#.

#:." From right-"Delta CMB, CB^2=CM^2+MB^2,#

# i.e., r^2=5^2+12^2=13^2. :. r=13#,

Thus, #S" has radius "r=13," and centre "C(5,+-13)#.

#:. S : (x-5)^2+(y+-13)^2=13^2#,

# or, S : x^2+y^2-10x+-26y+25=0#.

graph{(x^2+y^2-10x+26y+25)(x^2+y^2-10x-26y+25)=0 [-58.5, 58.5, -29.26, 29.3]}

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