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Mathematics Question on Coordinate Geometry

If the image of the point (−4,5)(-4, 5) in the line x+2y=2x + 2y = 2 lies on the circle (x+4)2+(y−3)2=r2(x + 4)^2 + (y - 3)^2 = r^2, then rr is equal to:

A

1

B

2

C

75

D

3

Answer

2

Explanation

Solution

The equation of the line is:

x+2y−2=0.x + 2y - 2 = 0.
The image of a point (x1,y1)(x_1, y_1) in a line ax+by+c=0ax + by + c = 0 is given by:

x−x1a=y−y1b=−2×ax1+by1+ca2+b2.\frac{x - x_1}{a} = \frac{y - y_1}{b} = -2 \times \frac{ax_1 + by_1 + c}{a^2 + b^2}.
Substitute (x1,y1)=(−4,5)(x_1, y_1) = (-4, 5), a=1a = 1, b=2b = 2, c=−2c = -2:

x+41=y−52=−2×1(−4)+2(5)−212+22.\frac{x + 4}{1} = \frac{y - 5}{2} = -2 \times \frac{1(-4) + 2(5) - 2}{1^2 + 2^2}.
Simplify:

x+41=y−52=−2×−4+10−21+4=−2×45.\frac{x + 4}{1} = \frac{y - 5}{2} = -2 \times \frac{-4 + 10 - 2}{1 + 4} = -2 \times \frac{4}{5}.
Solve for xx and yy:

x+4=−85  ⟹  x=−4−85=−285.x + 4 = -\frac{8}{5} \implies x = -4 - \frac{8}{5} = -\frac{28}{5}. y−5=−165  ⟹  y=5−165=255−165=95.y - 5 = -\frac{16}{5} \implies y = 5 - \frac{16}{5} = \frac{25}{5} - \frac{16}{5} = \frac{9}{5}.
The image of (−4,5)(-4, 5) is (−285,95)\left( -\frac{28}{5}, \frac{9}{5} \right).
Substitute this point into the circle equation (x+4)2+(y−3)2=r2(x + 4)^2 + (y - 3)^2 = r^2: (−285+4)2+(95−3)2=r2.\left( -\frac{28}{5} + 4 \right)^2 + \left( \frac{9}{5} - 3 \right)^2 = r^2.
Simplify each term:

−285+4=−285+205=−85,-\frac{28}{5} + 4 = -\frac{28}{5} + \frac{20}{5} = -\frac{8}{5}, 95−3=95−155=−65.\frac{9}{5} - 3 = \frac{9}{5} - \frac{15}{5} = -\frac{6}{5}.
Substitute:

(−85)2+(−65)2=r2.\left( -\frac{8}{5} \right)^2 + \left( -\frac{6}{5} \right)^2 = r^2.
Simplify:

6425+3625=r2  ⟹  10025=r2  ⟹  r2=4.\frac{64}{25} + \frac{36}{25} = r^2 \implies \frac{100}{25} = r^2 \implies r^2 = 4.
Thus: r=2.r = 2.