Question

Let $y=x+2,4 y=3 x+6$ and $3 y=4 x+1$ be three tangent lines to the circle $(x-h)^2+(y- k )^2= r ^2$ Then $h + k$ is equal to :

• A. 5
• B. $5(1+\sqrt{2})$
• C. 6
• D. $5 \sqrt{2}$

Answer 1

The Correct Option is A

To solve the problem of finding \(h + k\) for the given tangent lines to the circle, we will use the concept that the perpendicular distance from the center of the circle to a tangent is equal to the radius of the circle. The equations of the tangent lines provided are:

• Line 1: \(y = x + 2\)
• Line 2: \(4y = 3x + 6\)
• Line 3: \(3y = 4x + 1\)

We first convert the equations into a standard line form \(Ax + By + C = 0\):

• Line 1: \(x - y + 2 = 0\)
• Line 2: \(3x - 4y + 6 = 0\)
• Line 3: \(4x - 3y + 1 = 0\)

Given that each line is tangent to the circle \((x - h)^2 + (y - k)^2 = r^2\), the perpendicular distance from the center \((h, k)\) to each line equals \(r\).

Using the distance formula for a point \((h, k)\) to a line \(Ax + By + C = 0\), given as:

\(\text{Distance} = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}\)

This distance should be equal to the radius \(r\) for each tangent line.

Calculating the distances:

1. For Line 1: \(\frac{|h - k + 2|}{\sqrt{1^2 + (-1)^2}} = \frac{|h - k + 2|}{\sqrt{2}}\)
2. For Line 2: \(\frac{|3h - 4k + 6|}{\sqrt{3^2 + (-4)^2}} = \frac{|3h - 4k + 6|}{5}\)
3. For Line 3: \(\frac{|4h - 3k + 1|}{\sqrt{4^2 + (-3)^2}} = \frac{|4h - 3k + 1|}{5}\)

All these should equal \(r\), implying:

1. \(|h - k + 2| = r\sqrt{2}\)
2. \(|3h - 4k + 6| = 5r\)
3. \(|4h - 3k + 1| = 5r\)

Therefore \(h + k = 5\).

The correct option is 5.