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 :
- 5
- $5(1+\sqrt{2})$
- 6
- $5 \sqrt{2}$
The Correct Option is A
Approach Solution - 1
We are given three tangent lines to the circle, and we need to find the center of the circle, denoted by \((h, k)\).
Step 1: Equations of the Lines
The three lines are given as:
- \(L_1: y = x + 2\),
- \(L_2: 4y = 3x + 6 \implies y = \frac{3}{4}x + \frac{3}{2}\),
- \(L_3: 3y = 4x + 1 \implies y = \frac{4}{3}x + \frac{1}{3}\).
Step 2: Angle Bisector of \(L_1\) and \(L_2\)
The center of the circle lies on the angle bisector of the lines \(L_1\) and \(L_2\). Using the formula for the angle bisector of two lines, we get:
\[ \frac{4x - 3y + 1}{\sqrt{4^2 + (-3)^2}} = \pm \frac{3x - 4y + 6}{\sqrt{3^2 + (-4)^2}}. \]
Simplify the equation by considering the positive case:
\[ 4x - 3y + 1 = 3x - 4y + 6. \]
Rearrange the terms:
\[ x + y = 5. \]
Thus, the center lies on the line \(x + y = 5\), which is the angle bisector of \(L_1\) and \(L_2\).
Step 3: Angle Bisector of \(L_2\) and \(L_3\)
The center also lies on the angle bisector of \(L_2\) and \(L_3\). The equation of this bisector is:
\[ 3x - 4y + 6 = 0. \]
Step 4: Solve the System of Equations
We solve the system of equations:
- \(x + y = 5\),
- \(3x - 4y + 6 = 0\).
From the second equation, solve for \(x\):
\[ 3x = 4y - 6 \implies x = \frac{4y - 6}{3}. \]
Substitute \(x = \frac{4y - 6}{3}\) into \(x + y = 5\):
\[ \frac{4y - 6}{3} + y = 5. \]
Multiply through by 3 to eliminate the fraction:
\[ 4y - 6 + 3y = 15 \implies 7y = 21 \implies y = 3. \]
Substitute \(y = 3\) into \(x + y = 5\):
\[ x + 3 = 5 \implies x = 2. \]
Step 5: Center of the Circle
The center of the circle is \((h, k) = (2, 3)\). Thus:
\[ h + k = 2 + 3 = 5. \]
Conclusion
The value of \(h + k\) is:
\[ \boxed{5}. \]
Approach Solution -2
Bisector of lines
Centre lies on Bisector of
(0)
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Concepts Used:
Circle
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
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