Question

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?

Answer 1

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 )`

Answer 2

`(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.!