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CONIC SECTIONS — MATHS STD 11 Science — Question

Gujarat 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 $(x-h)^2+(y-k)^2=r^2 \ldots(i)$
Since, circle passes through the points $(2,-2)$ and $(3,4)$, so the points $(2,-2)$ and $(3,4)$ will lie on Eq. (i).
$\therefore(2-h)^2+(-2-k)^2=r^2 \ldots \text { (iii) }$
$\text { and }(3-h)^2+(4-k)^2=r^2 \ldots \text { (iii) }$
Now, from Eqs. (ii) and (iii), we get
$(2-h)^2+(-2-k)^2=(3-h)^2+(4-k)^2$
$\Rightarrow 4+h^2-4 h+4+k^2+4 k=9+h^2-6 h+16+k^2-8 k$
$\Rightarrow 2 h+12 k=17 \ldots \text { (iv) }$
Also, given that centre ( $h, k$ ) lies on $x+y=2$. So, it will satisfy it.
$\therefore h+k=2 \ldots(v)$
On solving Eqs. (iv) and (v), we get
$\mathrm{h}=0.7, \mathrm{k}=1.3$
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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