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The equation of a circle is (x - 3)^2 + (y + 2)^2 = 25. The point (8, -2) is on the circle.

What is the equation of the line that is tangent to the circle at (8, -2)?
y = 8
x = 8
x = 3
y = 3

Respuesta :

tonb
From the equation, you can see that (3,-2) is the center of the circle (the terms are 0 for those x,y values). The mentioned point has the same y value as the center. That's good news, because it tells us that the tangent line is exactly vertical. The x coordinate is given as x=8. Vertical lines have an equation of x=... so x=8 is the right answer.

Answer:  the correct option is

(B) x = 8.

Step-by-step explanation:  The given equation of a circle is :

[tex](x-3)^2+(y+2)^2=25~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

The point (8, -2) lies on the circle.

We are to select the equation of the line that is tangent to the circle at (8, -2).

We know that

the radius of a circle and the tangent to the circle at the same point of contact are perpendicular to each other.

The STANDARD equation of a circle with center at the point (h, k) and radius of length r units is given by

[tex](x-h)^2+(y-k)^2=r^2.[/tex]

From equation (i), we have

[tex](x-3)^2+(y+2)^2=25\\\\\Rightarrow (x-3)^2+(y-(-2))^2=5^2.[/tex]

So, the center of the circle (i) is (3, -2).

Since the radius of the circle passes through the point (8, -2) and the center (3, -2), so the slope of the radius will be

[tex]m=\dfrac{-2-(-2)}{3-8}=\dfrac{0}{-5}=0.[/tex]

So, equation of the radius passing through the point (8, -2) will be

[tex]y-(-2)=m(x-8)\\\\\rightarrow y+2=0\\\\\Rightarrow y=-2.[/tex]

So, the tangent at the point (8, -2) will be of the form

[tex]x=k.[/tex]

Since the tangent passes through the point (8, -2), so we get

[tex]x=8.[/tex]

Thus, the required equation of the tangent line at the point (8, -2) is x = 8.

Option (B) is CORRECT.

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