For the circle $C$ with the equation $x^2+y^2-16 x-12 y+64=0$ match the List I with the List II given below.…

For the circle $C$ with the equation $x^2+y^2-16 x-12 y+64=0$ match the List I with the List II given below.
The correct match is $\begin{array}{llll}(\mathrm{i}) & (\mathrm{ii}) & (\mathrm{iii}) & (\mathrm{iv})\end{array}$
  1. $\begin{array}{llll}(\mathrm{D}) & (\mathrm{B}) & (\mathrm{A}) & (\mathrm{E})\end{array}$
  2. $\begin{array}{llll}(\mathrm{D}) & (\mathrm{A}) & (\mathrm{B}) & (\mathrm{E})\end{array}$
  3. $\begin{array}{llll}(\mathrm{C}) & (\mathrm{D}) & (\mathrm{A}) & (\mathrm{B})\end{array}$
  4. $\begin{array}{llll}(\mathrm{C}) & (\mathrm{E}) & (\mathrm{B}) & (\mathrm{A})\end{array}$

Solution

Given equation of circle is $ C=x^2+y^2-16 x-12 y+64=0 $ (i) Equation of polar at $(-5,1)$ w.r.t. to $C$ is $ \begin{array}{cc} x(-5)+y(1)-8(x-5)-6(y+1)+64=0 \\ \Rightarrow & -5 x+y-8 x+40-6 y-6+64=0 \\ \Rightarrow & -13 x-5 y+98=0 \\ \Rightarrow & 13 x+5 y=98 \end{array} $ (ii) On differentiating Eq. (i) w.r.t. to $x$, we get $ \begin{array}{rrrr} & 2 x+2 y \frac{d y}{d x}-16-12 \frac{d y}{d x}+0=0 \\ \Rightarrow & (2 y-12) \frac{d y}{d x}=(16-2 x) \\ \Rightarrow & \frac{d y}{d x}=\left(\frac{8-x}{y-6}\right) \end{array} $ $ \text { At }(8,0) \quad\left(\frac{d y}{d x}\right)_{(8,0)}=\frac{8-8}{0-6}=0 $ $\therefore$ Equation of tangent at $(8,0)$ is $(y-0)=0(x-8)$ $ \Rightarrow \quad y=0 $ (iii) Slope of normal is $ \begin{aligned} \frac{d y}{d x} & =\frac{6-y}{8-x} \\ \left(\frac{d y}{d x}\right)_{(2,6)} & =\frac{6-6}{8-2}=0 \end{aligned} $ Equation of normal is $ \begin{aligned} & & (y-6) & =0(x-2) \\ \Rightarrow & & y & =6 \end{aligned} $ $ \Rightarrow \quad y=6 $ (iv) Equation of the diameter of circle through $(8,12)$ is $ x=8 $

Asked in: AP EAMCET 2013

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