Conic Sections Question 214
Question: The equation of the image of circle $ x^{2}+y^{2}+16x-24y+183=0 $ by the line mirror $ 4x+7y+13=0 $ is
Options:
A) $ x^{2}+y^{2}+32x-4y+235=0 $
B) $ x^{2}+y^{2}+32x+4y-235=0 $
C) $ x^{2}+y^{2}+32x-4y-235=0 $
D) $ x^{2}+y^{2}+32x+4y+235=0 $
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Answer:
Correct Answer: D
Solution:
[d] The centre of the given circle is $ (-8,12) $ and radius is 5. The image of the circle will have the same radius, i.e. the radius of the required circle is 5. The centre D of the required circle is the image of the centre C of the given circle in the line mirror. If D be $ (\alpha ,\beta ) $ then $ \frac{\alpha +8}{4}=\frac{\beta -12}{7}=-2[ \frac{4\times -8+7\times 12+13}{4^{2}+7^{2}} ] $ [See straight line] Or, $ \frac{\alpha +8}{4}=\frac{\beta -12}{7}=\frac{-2\times 65}{65}=-2 $
$ \therefore \alpha =-16,\beta =-2 $
$ \therefore $ Required circle is $ {{(x+16)}^{2}}+{{(y+2)}^{2}}=5^{2} $
BETA
