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) \[{{x}^{2}}+{{y}^{2}}+32x+4y+235=0\]
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 \[\frac{\alpha +8}{4}=\frac{\beta -12}{7}=-2\left[ \frac{4\times -8+7\times 12+13}{{{4}^{2}}+{{7}^{2}}} \right]\] [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 \[{{(x+16)}^{2}}+{{(y+2)}^{2}}={{5}^{2}}\]
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