Find the equation of a line passing through A (4,0) and tangent to the circle + = 4. please explain everything, thank youuu

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

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?