Find the equation to the tangent onto the circle x² + y² = 25 which passes through a point (10, 10).?

There are two tangent lines through any exterior point.

This is the equation of the circle having a diameter with endpoints (0, 0) and (10, 10).

x² + y² - 10x - 10y = 0

Subtract that from the given equation.

x² + y² = 25

x² + y² - 10x - 10y = 0 (-)

--------------------------------

10x + 10y = 25

2x + 2y = 5

That is the equation of the common chord of the two circles. It meets the given circle at the points of tangency. Solve this system of equations for the points of tangency:

{ x² + y² = 25

{ 2x + 2y = 5

After solving that you will have two points on each of the tangent lines. That should be all you need.

Followup:

While I was supposed to be doing something more important I went back to this problem using a different approach.

Let m be the slope of one of the tangent lines. Point (10, 10) is on the line, so the equation can be written like this:

y = mx - 10m + 10

Substitute that for y in the given circle equation. The result is this quadratic equation in x:

x² + (mx - 10m + 10)² = 25

(1 + m²)x² + (20m - 20m²)x + (100m² - 200m + 75) = 0

The tangent line meets the circle at only one point, so the quadratic equation has exactly one solution. That makes the discriminant zero.

(20m - 20m²)² - 4(1 + m²)(100m² - 200m + 75) = 0

Don't let the quartic equation be daunting. The higher terms disappear when it is expanded. It is then simplified into this quadratic equation in m:

3m² - 8m + 3 = 0

That results in two possible slopes, just as I promised.

m = [8 ± √(7)]/3

let (p, q) be the point where tangent touches the circle.

slope of tangent (ie line passing through (p, q) and (10, 10)

= (10 - q) / (10 - p) .. .. .. (1)

now we find slope of curve (circle x2 + y2 = 25)

2x + 2y dy/dx = 0

dy/dx = - x/y

slope of curve at (p, q)

= -p/q .. .. .. (2)

slope (1) and slope (2) must be equal

(10 - q) / (10 - p) = -p/q

p2 + q2 = 10(p + q) .. .. .. (3)

since (p, q) lies on curve (circle x2 + y2 = 25)

p2 + q2 = 25 .. .. .. (4)

consider eqn (3) and (4). we solve for p and q

10(p + q) = 25

p + q = 2.5 .. .. .. (5)

(p + q)2 = p2 + q2 + 2pq

2.5^2 = 25 + 2pq

2pq = 2(-25 + 6.25) = -18.75

(p - q)2 = p2 + q2 - 2pq = 25 + 18.75 = 43.75

p - q = sqrt(47.5) and p - q = -sqrt(47.5) .. .. .. (6)

we have two pairs of equations to solve for p and q:

p + q = 2.5

p - q = 6.9 .. .. .. / sqrt(47.5) = 6.9

and

p + q = 2.5

p - q = -6.9

first pair of equations:

p = [2.5 + 6.9 / 2

q = [2.5 - 6.9] / 2

p = 4.7

q = -2.2

second pair of equations:

p = [2.5 - 6.9] / 2

q = [2.5 + 6.9] / 2

p = -2.2

q = 4.7

so we have two points (4.7, -2.2) and (-2.2, 4.7) on the circle where line can touch. thus we have two tangents

tangent 1, passes through (10, 10) and (4.7, -2.2)

tangent 2, passes through (10, 10) and (-4.7, 2.2)

equations not worked out but can be easily found from above.

Differentiating implicitly w.r.t. x we get,

2x + 2y(dy/dx) = 0

=> dy/dx = -x/y

so, passing through (10, 10) we have dy/dx = -1...i.e. gradient.

Then, equation is y - 10 = -(x - 10)

=> y = 20 - x

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