Circle 4 Question 16
16. The points of intersection of the line $4 x-3 y-10=0$ and the circle $x^{2}+y^{2}-2 x+4 y-20=0$ are…and… .
$(1983,2 M)$
Analytical & Descriptive Questions
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Answer:
Correct Answer: 16. $(-2,-6)$ and $(4,2)$
Solution:
- For point of intersection, we put
$$ \begin{array}{lc} & x=\frac{3 y+10}{4} \text { in } x^{2}+y^{2}-2 x+4 y-20=0 \\ & \\ \Rightarrow & \frac{3 y+10^{2}}{4}+y^{2}-2 \frac{3 y+10}{4}+4 y-20=0 \\ \Rightarrow & 25 y^{2}+100 y-300=0 \\ \Rightarrow & y^{2}+4 y-12=0 \\ \Rightarrow & (y-2)(y+6)=0 \\ \Rightarrow & y=-6,2 \\ \text { When } & y=-6 \Rightarrow x=-2 \\ \text { When } & y=2 \\ \Rightarrow & x=4 \end{array} $$
$\therefore$ Point of intersection are $(-2,-6)$ and $(4,2)$.
