- Home
- Standard 11
- Mathematics
The line $L$ passes through the points of intersection of the circles ${x^2} + {y^2} = 25$ and ${x^2} + {y^2} - 8x + 7 = 0$. The length of perpendicular from centre of second circle onto the line $L$, is
$4$
$3$
$1$
$0$
Solution
(d) Let $AB$ be the line of intersection of the two circles
${x^2} + {y^2} = 25$….$(i)$
${x^2} + {y^2} – 8x + 7 = 0$….$(ii)$
Solving $(i)$ and $(ii),$ we get coordinates of $A$ and $B$.
Subtracting $(i)$ from $(ii),$ we get
$ – 8x + 32 = 0$ or $x = 4$
Hence from $(i),$ we get $16 + {y^2} = 25$
$\therefore $${y^2} = 9$ or $y = \pm 3$.
Thus coordinates of $A$ and $B$ are $(4, 3)$ and $(4, -3)$.
Hence equation of $L$ is $\frac{{y – 3}}{{3 + 3}} = \frac{{x – 4}}{{4 – 4}}$ or $x – 4 = 0$.
Also coordinates of centre $C$ of second circle is $(4, 0)$.
Hence $CM = $ Length of perpendicular from $C$ to the line $L = \frac{{4 – 4}}{{\sqrt 1 }} = 0$.
So, $C$ and $M$ are the same point.
