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  • The equation of the image of the circle <img alt="\mathrm{x^2+y^2+16 x-24 y+183=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&plus;16%20x-24%20y&

The equation of the image of the circle \mathrm{x^2+y^2+16 x-24 y+183=0} by the line mirror \mathrm{4 x+7 y+13=0} is

Option: 1

\mathrm{x^2+y^2+32 x-4 y+235=0}


Option: 2

\mathrm{x^2+y^2+32 x+4 y-235=0}


Option: 3

\mathrm{x^2+y^2+32 x-4 y-235=0}


Option: 4

\mathrm{x^2+y^2+32 x+4 y+235=0}


Answers (1)

best_answer

The given circle and line are \mathrm{x^2+y^2+16 x-24 y+183=0}       \mathrm{...(i)}

and \mathrm{4 x+7 y+13=0}                                    \mathrm{...(ii)}

Centre and radius of circle (i) are (-8,12) and 5 respectively. Let the centre of the image circle be \mathrm{ \left(x_1, y_1\right)}

Then slope of \mathrm{ C_1 C_2 \times} slope of \mathrm{ 4 x+7 y+13=-1}                                           

\mathrm{ \Rightarrow\left(\frac{y_1-12}{x_1+8}\right) \times\left(-\frac{4}{7}\right)=-1 \quad \text { or } 4 y_1-48=7 x_1+56 }

or \mathrm{ 7 x_1-4 y_1+104=0}                                 \mathrm{ ...(iii)}

and mid point of \mathrm{ C_1 C_2 i.e., \left(\frac{x_1-8}{2}, \frac{y_1+12}{2}\right) lie \, \, on \, \, 4 x+7 y+13=0},

then \mathrm{ 4\left(\frac{x_1-8}{2}\right)+7\left(\frac{y_1+12}{2}\right)+13=0 \text { or } 4 x_1+7 y_1+78=0 }                    \mathrm{ ...(iv)}

Solving (iii) and (iv), we get \mathrm{\left(x_1, y_1\right)=(-16,-2)}

\therefore Equation of the image circle is \mathrm{(x+16)^2+(y+2)^2=5^2}   or   \mathrm{ x^2+y^2+32 x+4 y+235=0}

 

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