JEE Mathematics 2023 Question with Solution

Given: The lines $L_1: y = x + 2$, $L_2: 4y = 3x + 6$ and $L_3: 3y = 4x + 1$ are tangent to the circle $(x - h)^2 + (y - k)^2 = r^2$.

Find: The value of $h + k$.

The center of a circle tangent to two intersecting lines lies on an angle bisector of those lines.

Write the lines in standard form:

Using the angle bisector of $L_2$ and $L_3$ as shown in the solution,

Since both denominators are $5$, take the positive case:

So,

The center also lies on the other bisector used in the solution:

Now solve the system

From

we get

Substitute into $x + y = 5$:

Thus,

Hence,

Therefore the center is $(h,k) = (2,3)$, so

The correct option is A.

Given: The three tangent lines are $y = x + 2$, $4y = 3x + 6$ and $3y = 4x + 1$.

Find: $h + k$.

From the angle-bisector relation used in the solution, the center lies on

Since the center is $(h,k)$, it immediately follows that

So the correct option is A.