If the point of intersection of the pair of the transverse common tangents and that of the pair of direct…

If the point of intersection of the pair of the transverse common tangents and that of the pair of direct common tangents drawn to the circles $x^2+y^2-14 x+6 y+33=0$ and $x^2+y^2+30 x-2 y+1=0$ are $T$ and $D$ respectively, then the centre of the circle having TD as diameter is
  1. $\left(\frac{39}{2}, \frac{-7}{4}\right)$
  2. $\left(\frac{39}{4}, \frac{7}{2}\right)$
  3. $\left(\frac{39}{4}, \frac{-7}{2}\right)$
  4. $\left(\frac{39}{2}, \frac{-7}{2}\right)$

Solution


Given, $C_1: x^2+y^2+30 x-2 y+1=0$ $ \begin{aligned} & \therefore \text { centre }(O)=(-15,1) \\ & \text { and radius }=\sqrt{225+1-1}=\sqrt{225}=15 \\ & \text { and } C_2: x^2+y^2-14 x+6 y+33=0 \\ & \therefore \text { Centre }\left(O^{\prime}\right)=(7,-3) \\ & \text { and radius }=\sqrt{49+9-33}=\sqrt{25}=5 \end{aligned} $ Since, point $T$ divides $O O^{\prime}$ in $15: 5$ i.e., $3: 1$ internally. $ \therefore \quad T=\left(\frac{21-15}{4}, \frac{-9+1}{4}\right)=\left(\frac{3}{2},-2\right) $ Also, point $D$ divides $O O^{\prime}$ in $15: 5$ i.e., $3: 1$ externally. $ D=\left(\frac{21+15}{2}, \frac{-9-1}{2}\right)=(18,-5) $ Now, centre of circle with $T D$ as diameters of mid-point of $T D$ $ =\left(\frac{18+3 / 2}{2}, \frac{-2-5}{2}\right)=\left(\frac{39}{4}, \frac{-7}{2}\right) $

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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