Q99 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
The points of intersection of circles $(x+1)^2+y^2=4$ and $(x-1)^2+y^2=9$ are $(a, \pm b)$, then $(a, b)$ equals to
- A$(-1,2)$
- B$(1,3)$
- C$\left(1.25, \dfrac{3}{4} \sqrt{7}\right)$
- Correct answerD$\left(-1.25, \dfrac{3}{4} \sqrt{7}\right)$
Solution
The given equations of the circles are:
$(x+1)^2 + y^2 = 4$ (1)
$(x-1)^2 + y^2 = 9$ (2)
Expanding both equations:
$x^2 + 2x + 1 + y^2 = 4 \Rightarrow x^2 + y^2 + 2x = 3$ (3)
$x^2 - 2x + 1 + y^2 = 9 \Rightarrow x^2 + y^2 - 2x = 8$ (4)
Subtracting equation (4) from equation (3):
$(x^2 + y^2 + 2x) - (x^2 + y^2 - 2x) = 3 - 8$
$4x = -5$
$x = -1.25$
Substituting $x = -1.25$ into equation (3):
$(-1.25)^2 + y^2 + 2(-1.25) = 3$
$1.5625 + y^2 - 2.5 = 3$
$y^2 - 0.9375 = 3$
$y^2 = 3.9375$
$y^2 = \dfrac{39375}{10000} = \dfrac{63}{16}$
$y = \pm \sqrt{\dfrac{63}{16}} = \pm \dfrac{3\sqrt{7}}{4}$
Thus, the points of intersection are $(-1.25, \pm \dfrac{3}{4}\sqrt{7})$. Comparing this with $(a, \pm b)$, we get $a = -1.25$ and $b = \dfrac{3}{4}\sqrt{7}$.
Answer: $\left(-1.25, \dfrac{3}{4} \sqrt{7}\right)$
$(x+1)^2 + y^2 = 4$ (1)
$(x-1)^2 + y^2 = 9$ (2)
Expanding both equations:
$x^2 + 2x + 1 + y^2 = 4 \Rightarrow x^2 + y^2 + 2x = 3$ (3)
$x^2 - 2x + 1 + y^2 = 9 \Rightarrow x^2 + y^2 - 2x = 8$ (4)
Subtracting equation (4) from equation (3):
$(x^2 + y^2 + 2x) - (x^2 + y^2 - 2x) = 3 - 8$
$4x = -5$
$x = -1.25$
Substituting $x = -1.25$ into equation (3):
$(-1.25)^2 + y^2 + 2(-1.25) = 3$
$1.5625 + y^2 - 2.5 = 3$
$y^2 - 0.9375 = 3$
$y^2 = 3.9375$
$y^2 = \dfrac{39375}{10000} = \dfrac{63}{16}$
$y = \pm \sqrt{\dfrac{63}{16}} = \pm \dfrac{3\sqrt{7}}{4}$
Thus, the points of intersection are $(-1.25, \pm \dfrac{3}{4}\sqrt{7})$. Comparing this with $(a, \pm b)$, we get $a = -1.25$ and $b = \dfrac{3}{4}\sqrt{7}$.
Answer: $\left(-1.25, \dfrac{3}{4} \sqrt{7}\right)$
Hence, the correct answer is option: 'D'.