SATHEE: Circle 4 Question 16

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

$\begin{array}{lc} x = \frac{3 y + 10}{4} in x^{2} + y^{2} - 2 x + 4 y - 20 = 0 \\ \Rightarrow & \frac{3 y + 10^{2}}{4} + y^{2} - 2 \frac{3 y + 10}{4} + 4 y - 20 = 0 \\ \Rightarrow & 25 y^{2} + 100 y - 300 = 0 \\ \Rightarrow & y^{2} + 4 y - 12 = 0 \\ \Rightarrow & (y - 2) (y + 6) = 0 \\ \Rightarrow & y = - 6, 2 \\ When & y = - 6 \Rightarrow x = - 2 \\ When & y = 2 \\ \Rightarrow & x = 4 \end{array}$

$∴$ Point of intersection are $(- 2, - 6)$ and $(4, 2)$ .

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Circle 4 Question 16

16. The points of intersection of the line 4x−3y−10=0 and the circle x2+y2−2x+4y−20=0 are…and… .

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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… .