Gujarati
10-1.Circle and System of Circles
medium

The points of intersection of the circles ${x^2} + {y^2} = 25$and ${x^2} + {y^2} - 8x + 7 = 0$ are

A

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

B

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

C

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

D

$(4, 3) $ and $(3, 4)$ 

Solution

(a) ${x^2} + {y^2} = 25$ .….$(i)$

${x^2} + {y^2} – 8x + 7 = 0$ .….$(ii)$

From $(i)$ and $(ii),$ we get

$8x = 7 + 25$ or $x = 4$ and for $x = 4$, we get $y = \pm 3$.

Hence points of intersection are $(4, 3), (4, -3).$

Standard 11
Mathematics

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