Question

Circle O has center (2, 6) and passes through the point P (4, 3). Write the equation of the line that is tangent to circle O at point P
A:y=3/2,x-3
B:y=2/3,x+1/3
C:y=-3/2,x+9
D:y=-2/3x+17/3

Answer 1

First, fine the slope of the lines.

m=3-6/4-2
m=-3/2

To find the perpendicular slope of this, find the opposite reciprocal, or 2/3

Use formula y=mx+b to find the equation of this line.

3=(2/3)(4)+b
3=b+8/3
3-8/3=b+8/3-8/3
1/3=b

Now, plug in the values

y=2/3x+1/3, or B

Answer 2

The option (B) $y=(\frac{2}{3} )x+(\frac{1}{3} )$  is the correct answer.

We have ,

Center of the circle, $O(x1,y1)$ ⇒ $O(2,6)$

The point is $P(x2,y2)$ ⇒ $P(4,3)$

The equation of line that is tangent to circle and point can be find by using the slope formula and equation of line.

Formula for slope, $m=\frac{y2-y1}{x2-x1}$

On substituting the points,

$m=\frac{3-6}{4-2}$

⇒$m=\frac{-3}{2}$

The slope of the tangent is perpendicular to the slope of the line that passes through centre and on point of the circle.

So, the slope of tangent = $-\frac{1}{m}$

⇒$\frac{2}{3}$

Formula for equation of line, $y=mx+b$

Now we need to find $b$ at $P(4,3)$

⇒ $3= (\frac{2}{3})4+b$

⇒$3=(\frac{8}{3})+b$

⇒$3-(\frac{8}{3} )=b$

⇒$\frac{1}{3} =b$

Thus the equation of line, $y=mx+b$

⇒ $y=(\frac{2}{3} )x+(\frac{1}{3} )$

Hence, the option (B) is the correct answer.

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