Question

Find the equation of a line passing through A (4,0) and tangent to the circle "TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex-formula"> + $y^{2}$ = 4. please explain everything, thank youuu

Answer 1

Answer:

y = -1/√3 (x − 4)

Step-by-step explanation:

Let's say the point where the tangent line intersects the circle is B (bₓ, bᵧ).  Therefore:

bₓ² + bᵧ² = 4

The slope of the tangent line at that point is dy/dx, which we can find with implicit differentiation:

x² + y² = 4

2x + 2y dy/dx = 0

2y dy/dx = -2x

dy/dx = -x/y

At point B, the slope of the tangent line is m = -bₓ/bᵧ.

We can also write that as the slope between the points A(4, 0) and B(bₓ, bᵧ).

m = (bᵧ − 0) / (bₓ − 4)

m = bᵧ / (bₓ − 4)

Setting the expressions equal:

-bₓ / bᵧ = bᵧ / (bₓ − 4)

bᵧ² = -bₓ (bₓ − 4)

bᵧ² = 4 bₓ − bₓ²

Substituting this into the first equation:

bₓ² + (4 bₓ − bₓ²) = 4

4 bₓ = 4

bₓ = 1

Therefore, bᵧ = √3.  So the slope of the line is -1/√3.  The equation of the line is:

y − 0 = -1/√3 (x − 4)

y = -1/√3 (x − 4)

Graph: desmos.com/calculator/sornb1ajgp