The equation of tangent to the circle x^2+y^2=r^2 is y=mx+c ,then the point of contact of the circle and the tangent is?

2 Answers
Somebody N.
Sep 8, 2017

See below.

Explanation:

The point of contact can be found by equating the equations together. This will find a value that is common to both equations and consequently will be the point or points of contact.

So we have #x^2 + y^2 = r^2#

rearranging: #y =sqrt( r^2 -x^2)#

Tangent line: #y = mx + c #

Point of contact: #mx + c = sqrt(r^2 - x^2 )#

Ratnaker Mehta
Sep 9, 2017

#(1) :"The tgt. is, "y=mx+-rsqrt(m^2+1).#

#(2) :"The Point of contact is, "((-mc)/sqrt(m^2+1), c/sqrt(m^2+1)),#

where, #c=pmsqrt(m^2+!).#

Explanation:

Let, the Line # t : y=mx+c# be tangent (tgt.) to the Circle

# S : x^2+y^2=r^2.#

Clearly, #O(0,0)# is the Centre and #r# is the Radius of #S#.

From Geometry, we know that, in a circle, the #bot-#dist. from

the centre to a tgt. equals the radius.

Recall that, the #bot-#dist. from pt.

#(h,k)" to line "ax+by+c=0,# is #|ah+bk+c|/sqrt(a^2+b^2).#

Utilising these facts, we have,

#|m(0)-0+c|/sqrt(m^2+1)=r, or, c=+-rsqrt(m^2+1)..........(ast).#

Hence, the tgt. #t# is, #y=mx+-rsqrt(m^2+1).#

If #P# is the pt. of contact, from Geometry, we know that,

#OP bot t, and, OPnn t={P}.#

So, #P# can be obtained by solvig the eqns. of #t" and, line "OP.#

To find the eqn. of #OP#, we note that,

its slope is #-1/m, &, O in OP.#

#:. OP :y-0=-1/m(x-0), i.e., y=-1/m*x............(1).#

# t : y=mx+c.........................................................(2).#

Solving #(1), and, (2), x=(-mc)/sqrt(m^2+1), y=c/sqrt(m^2+1).#

Therefore, the pt. of contact is,#((-mc)/sqrt(m^2+1), c/sqrt(m^2+1)),#

#"where, "c=pmrsqrt(m^2+1).#

Enjoy Maths.!

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