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Conic Sections — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsConic Sections2 Marks
Question
Find the equation of the circle which passes through the points $(2, - 2)$ and $(3, 4)$ and whose centre lies on the line $x + y = 2.$
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Answer

Let the equation of circle with centre $(h, k)$ and radius $r$ be $(x - h)^2 + (y - k)^2 = r^2 ...(i)$
Since, circle passes through the points $(2, - 2)$ and $(3, 4),$ so the points $(2, - 2)$ and $(3, 4)$ will lie on Eq. $(i).$
$\therefore (2 - h)^2 + (- 2 - k)^2 = r^2 ...(ii)$
and $(3 - h)^2 + (4 - k)^2 = r^2...(iii)$
Now, from Eqs. $(ii)$ and $(iii),$ we get
$(2 - h)^2 + (- 2 - k)^2 = (3 - h)^2 + (4 - k)^2$ 
$\Rightarrow 4 + h^2 - 4h + 4 + k^2 + 4k = 9 + h^2 - 6h + 16 + k^2- 8k$
$\Rightarrow 2h + 12k = 17 ...(iv)$
Also, given that centre $(h, k)$ lies on $x + y = 2.$ So, it will satisfy it.
$\therefore h + k = 2 ...(v)$
On solving Eqs. $(iv)$ and $(v)$, we get
$h = 0.7, k = 1.3$
Now, $r^2 = (2 - 0.7)^2 + (- 2 - 1.3)^2 = 1.69 + 10.89 = 12.58$
On putting $h = 0.7, k = 13$ and $r^2 = 12.58$ in Eq. (i), we get
$(x - 0.7)^2 + (y - 1.3)^2 = 12.58$
which is the required equation of circle.

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