Circle Ques 61
- Find the equation of circle touching the line $2 x+3 y+1=0$ at the point $(1,-1)$ and is orthogonal to the circle which has the line segment having end points $(0,-1)$ and $(-2,3)$ as the diameter.
(2004, 4M)
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Answer:
Correct Answer: 61.$2 x^{2}+2 y^{2}-10 x-5 y+1=0$ 21. $y=0$ and $7 y-24 x+16=0$
Solution:
Formula:
- The equation of circle having tangent $2 x+3 y+1=0$ at $(1,-1)$
$$ \begin{alignedat} & \Rightarrow \quad(x-1)^{2}+(y+1)^{2}+\lambda(2 x+3 y+1)=0 \\ & x^{2}+y^{2}+2 x(\lambda-1)+y(3 \lambda+2)+(\lambda+2)=0 \end{aligned} $$
which is orthogonal to the circle with endpoints of diameter $(0,-1)$ and $(-2,3)$.
$$ \begin{array}{lc} \Rightarrow & x(x+2)+(y+1)(y-3)=0 \\ \text { or } & x^{2}+y^{2}+2 x-2 y-3=0 \\ \therefore & \frac{2(2 \lambda-2)}{2} \cdot 1+\frac{2(3 \lambda+2)}{2}(-1)=\lambda+2-3 \\ \Rightarrow & 2\lambda - 2 - 3\lambda - 2 = \lambda - 1 \\ \Rightarrow \quad 2 \lambda=-3 \ \Rightarrow & \lambda=-3 / 2 \end{array} $$
From Eq. (i), the equation of a circle,
$$ 2 x^{2}+2 y^{2}-10 x-5 y+1=0 $$
BETA
