Intersection points of the circles x2+y2=25 and x2+y2-8x+7=0, are

Question

Intersection points of the circles x2+y2=25 and x2+y2-8x+7=0, are (A) (4, 3) and (4,-3)  (B)  (4,-3) and (-4,-3)  (C)  (-4,3) and (4, 3) (D) (4, 3) and (3, 4)

Tardigrade Admin · Accepted Answer

Given, x2+y2=25 ...(i) and x2+y2-8x+7=0 ...(ii) From Eqs. (i) and (ii), 25-8x+7=0 ⇒ 8x=32 ⇒ x=4 and y=± 3 Hence, intersection points are (4,3) and (4,-3) .

Q. Intersection points of the circles $x^{2} + y^{2} = 25$ and $x^{2} + y^{2} - 8 x + 7 = 0,$ are

$(4, 3)$ and $(4, - 3)$

$(4, - 3)$ and $(- 4, - 3)$

$(- 4, 3)$ and $(4, 3)$

$(4, 3)$ and $(3, 4)$

Given, $x^{2} + y^{2} = 25$ ...(i) and $x^{2} + y^{2} - 8 x + 7 = 0$ ...(ii)
From Eqs. (i) and (ii), $25 - 8 x + 7 = 0$ $\Rightarrow$ $8 x = 32$ $\Rightarrow$
$x = 4$ and $y = \pm 3$
Hence, intersection points are (4,3) and $(4, - 3)$ .

Q. Intersection points of the circles x2+y2=25 and x2+y2−8x+7=0, are

(4,3) and (4,−3)

(4,−3) and (−4,−3)

(−4,3) and (4,3)

(4,3) and (3,4)