SECTION – A PHYSICS If the image of the point (-4, 5) in the line… — Solution | JEE Mathematics

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SECTION – A PHYSICS

If the image of the point $(-4, 5)$ in the line $x + 2y = 2$ lies on the circle $(x + 4)^2 + (y - 3)^2 = r^2$ , then $r$ is equal to:

A
$1$
B
$2$
C
$75$
D
$3$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The point is $(-4, 5)$ , the line is $x + 2y - 2 = 0$ , and the circle is $(x + 4)^2 + (y - 3)^2 = r^2$ .

Find: The value of $r$ .

First find the image of the point $(-4, 5)$ in the line $x + 2y - 2 = 0$ . For reflection of a point $(x_0, y_0)$ in the line $ax + by + c = 0$ , the image $(x_1, y_1)$ is given by

x_1 = \frac{x_0(b^2 - a^2) - 2y_0ab - 2ac}{a^2 + b^2}

and

y_1 = \frac{y_0(a^2 - b^2) - 2x_0ab - 2bc}{a^2 + b^2}

Detailed Calculation

Here $a = 1$ , $b = 2$ , $c = -2$ , and $(x_0, y_0) = (-4, 5)$ .

So,

x_1 = \frac{-4(2^2 - 1^2) - 2 \times 5 \times 1 \times 2 - 2 \times 1 \times (-2)}{1^2 + 2^2}

= \frac{-4(3) - 20 + 4}{5}

= \frac{-12 - 20 + 4}{5} = \frac{-28}{5}

Also,

y_1 = \frac{5(1^2 - 2^2) - 2 \times (-4) \times 1 \times 2 - 2 \times 2 \times (-2)}{1^2 + 2^2}

= \frac{5(-3) + 16 + 8}{5}

= \frac{-15 + 24}{5} = \frac{9}{5}

Hence, the image point is

\left(-\frac{28}{5}, \frac{9}{5}\right)

Now substitute this point into the circle equation:

\left(-\frac{28}{5} + 4\right)^2 + \left(\frac{9}{5} - 3\right)^2 = r^2

Compute each term:

-\frac{28}{5} + 4 = -\frac{8}{5}

\frac{9}{5} - 3 = -\frac{6}{5}

Therefore,

\left(-\frac{8}{5}\right)^2 + \left(-\frac{6}{5}\right)^2 = r^2

\frac{64}{25} + \frac{36}{25} = r^2

\frac{100}{25} = r^2

4 = r^2

So,

r = 2

Therefore, the value of $r$ is $2$ . The correct option is B.

Using Reflection Formula Directly

Given: Reflect the point $(-4, 5)$ in the line $x + 2y - 2 = 0$ and use the circle equation.

Find: The value of $r$ .

Using the direct reflection relation

\frac{x - x_1}{a} = \frac{y - y_1}{b} = -2 \times \frac{ax_1 + by_1 + c}{a^2 + b^2}

with $a = 1$ , $b = 2$ , $c = -2$ , and $(x_1, y_1) = (-4, 5)$ ,

\frac{x + 4}{1} = \frac{y - 5}{2} = -2 \times \frac{1(-4) + 2(5) - 2}{1^2 + 2^2}

= -2 \times \frac{4}{5} = -\frac{8}{5}

Thus,

x + 4 = -\frac{8}{5} \Rightarrow x = -\frac{28}{5}

y - 5 = -\frac{16}{5} \Rightarrow y = \frac{9}{5}

So the reflected point is

\left(-\frac{28}{5}, \frac{9}{5}\right)

Substitute into the circle:

\left(-\frac{28}{5} + 4\right)^2 + \left(\frac{9}{5} - 3\right)^2 = r^2

\left(-\frac{8}{5}\right)^2 + \left(-\frac{6}{5}\right)^2 = r^2

\frac{64}{25} + \frac{36}{25} = 4 = r^2

Hence, $r = 2$ . The correct option is B.

Common mistakes

A common mistake is using the original point $(-4, 5)$ directly in the circle equation. This is wrong because the question asks for the image of the point in the line. First reflect the point, then substitute the reflected coordinates into the circle.
Students often write the line as $x + 2y = 2$ and forget to convert it into the standard form $ax + by + c = 0$ . This leads to incorrect values of $a$ , $b$ , and $c$ in the reflection formula. Use $x + 2y - 2 = 0$ .
Another mistake is concluding $r^2 = 4$ and marking the answer as $4$ . This is wrong because the circle equation gives the square of the radius. After finding $r^2$ , take the positive square root to get $r = 2$ .

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Step-by-step solution

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