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# A line is tangent to a circle

Question:

A circle has the equation x2 + (y - 1)2 = 8. A line is tangent to the circle at P (2, 3). Find the equation of the line that is tangent to the circle.

Solution:

Let Q be the center of the circle. From the circle equation x2 + (y - 1)2 = 8, we get the center of the circle is Q (0, 1) and the radius of the circle r = square root of 8 = square root of (2 × 4) = square root of (2 × 22) = 2 square root of 2.

We are given that a line is tangent to the circle at P (2, 3). So, the point P (2, 3) is the point of tangency. Name the line as line L. Point Q is the center of the circle. Point P is the point of the tangency. so, QP is the radius of the circle. By the circle property, a radius draw to a point of tangency is perpendicular to the tangent. So, the line L is perpendicular to the radius QP at the point P.

Let k be the slope, from slope definition kQP = (y2 - y1)/(x2 - x1)

Naming P is the second point, so, x2 = 2 and y2 = 3.

Naming Q is the first point, so, x1 = 0 and y1 = 1.

kQP = (y2 - y1)/(x2 - x1) = (3 - 1)/(2 - 0) = 2/2 = 1

Because the line L is perpendicular to QP, so, kL × kQP= -1

kL = (-1)/kQP = -1/1 = -1

Now, we get the slope of the line L is -1.

The line L has the equation y = kx + b, substitute k = -1, we get y = -x + b

Because the point P (2, 3) lies in the line L, so the point P satisfy the line equation. Substitute the point P (2, 3) into the line equation y = - x + b

3 = -2 + b -> b = 5.

So, the line equation is y = -x + 5 In the figure above, the blue curve is the circle x2 + (y - 1)2 = 8. The point P (2, 3) is the point of tangency and the line equation is y = -x + 5. The line is tangent to the circle at the point P. Watch the video for more details.