Question

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

• 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. $x^2+y^2+32x+4y+235=0$

Answer 1

The correct option is D $x^2+y^2+32x+4y+235=0$

Equation of given circle:

$x^2+y^2++16x-24y+183=0$

$\text{Idea}$; Lets find the reflection of centre of this circle with respect to the given line equation. Then find the radius of given circle. After reflection also, the radius of circle does not change. Thus finally knowing the centre of reflected circle and its radius we can find equation of reflected circle.

First convert this equation to standard form of circle equation as follows:

$x^2+y^2++16x-24y+183=0$

$(x^2+16x+64)+(y^2-24y+144)-25=0$

$(x+8{)}^2+(y-12{)}^2-25=0$

$(x+8{)}^2+(y-12{)}^2=5^2$

Thus $\text{Centre}=(-8,12)$

and $\text{radius}=\sqrt{25}=5$

Let reflected centre have co-ordinates= $(h,k)$

Now note that mid point of actual centre and reflected centre will definitely fall on line of reflection.

Mid-point of $(h,k)$ and $(-8,12)$=$(\frac{h-8}{2},\frac{k+12}{2})$

Lets try to satisfy line equation with given point.

$4x+7y+13=0\to 4\left(\frac{h-8}{2}\right)+7\left(\frac{k+12}{2}\right)+13=0$

$4h+7k+78=0$ .....(1)

Also slope of given line is perpendicular to slope of RC

Slope of given line$=-4/7$

$m_1{m}_2=-1$

$\frac{\frac{(k+12)}{2}-12}{\left(\frac{h-8}{2}\right)+8}\times \frac{-4}{7}=-1$

$\to 7h-4k+104=0$ ...........(2)

Solving (1) and (2) we get, $(h,k)=(-16,-2)$

Equation of circle with centre $(-16,-2)$ and radius $=5$:

$(x+16{)}^2+(y+2{)}^2=5^2$

$x^2+y^2+32x+4y+235=0$