Find the equations of the circle...

MathematicsStandard and General Equation of a CircleJEE Advanced 1982Moderate

Find the equations of the circle passing through $(- 4, 3)$ and touching the lines $x + y = 2$ and $x - y = 2$ .

Answer:x^2 + y^2 + 2(10 \pm \sqrt{54})x + 55 \pm \sqrt{54} = 0

Visualized Solution (Hindi)

Visualizing the Geometry

Given lines: $L_{1} : x + y - 2 = 0$ and $L_{2} : x - y - 2 = 0$ .
Point on circle: $P (- 4, 3)$ .
Intersection of $L_{1}$ and $L_{2}$ : $(2, 0)$ .

Identifying the Angle Bisector

The center must lie on the angle bisector of the lines.
Bisectors are $y = 0$ and $x = 2$ .
Given $P (- 4, 3)$ , the center must lie on the x-axis ( $y = 0$ ).
Let the center be $C (a, 0)$ .

Defining the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Radius $r$ is the distance from $(a, 0)$ to $x + y - 2 = 0$ .
$r = \frac{∣ a + 0 - 2∣}{1 ^{2} + 1 ^{2}} = \frac{∣ a - 2∣}{2}$
Equation of circle: $(x - a)^{2} + (y - 0)^{2} = r^{2}$

Applying the Point Constraint

Substitute $P (- 4, 3)$ into $(x - a)^{2} + y^{2} = \frac{( a - 2 ) ^{2}}{2}$ :
$(- 4 - a)^{2} + 3^{2} = \frac{( a - 2 ) ^{2}}{2}$
$(a + 4)^{2} + 9 = \frac{( a - 2 ) ^{2}}{2}$

Expanding the Equation

Expand: $2 (a^{2} + 8 a + 16 + 9) = a^{2} - 4 a + 4$
$2 (a^{2} + 8 a + 25) = a^{2} - 4 a + 4$
$2 a^{2} + 16 a + 50 = a^{2} - 4 a + 4$

Forming the Quadratic

Subtract $a^{2} - 4 a + 4$ from both sides:
$a^{2} + 20 a + 46 = 0$

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Using $a = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ :
$a = \frac{- 20 \pm 400 - 184}{2} = \frac{- 20 \pm 216}{2}$
$a = - 10 \pm 54$

The Final Equations

Substitute $a = - 10 \pm 54$ into the expanded form:
$x^{2} + y^{2} - 2 a x - 8 a - 25 = 0$
Final Equations:
$x^{2} + y^{2} + 2 (10 \pm 54) x + 55 \pm 854 = 0$

Key Takeaways

Key Takeaway: Center of a circle touching two lines lies on their angle bisector.
Symmetry: Use the position of the given point to choose the correct bisector.
Next Challenge: How would the equation change if the circle had a fixed radius $r$ instead of passing through a point?

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