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JEE MainMathematicsModerateCircles

Find the number of integer values of k for which the equation
x2+y2+(k−1)x−ky+5=0x^{2} + y^{2} + (k - 1)x - ky + 5 = 0 represents a circle whose radius
cannot exceed 3.
k ds mu iw.kk±dh; ekuksa Kkr dhft, ftuds fy, lehdj.k
x2+y2+(k−1)x−ky+5=0x^{2} + y^{2} + (k - 1)x - ky + 5 = 0,d,slk o`Ùk fu:fir djrk

Show answer and solution
x2+y2+(k−1)x−ky+5=0x^{2} + y^{2} + (k - 1)x - ky + 5 = 0
this represents a circle whose radius
R is (k−12)2+(k2)2−5R\ is\ \sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5}

R (k−12)2+(k2)2−5R\ \sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5}
So blfy,
(k−12)2+(k2)2−5≤3\sqrt{\left( \frac{k - 1}{2} \right)^{2} + \left( \frac{k}{2} \right)^{2} - 5} \leq 3
k2−2k+14+k24−5≤9\frac{k^{2} - 2k + 1}{4} + \frac{k^{2}}{4} - 5 \leq 9
0≤2k2−2k−19≤360 \leq 2k^{2} - 2k - 19 \leq 36
⇒k∈[1−1112,1−372]∪[1+372,1+1112]\Rightarrow k \in \left\lbrack \frac{1 - \sqrt{111}}{2},\frac{1 - \sqrt{37}}{2} \right\rbrack \cup \left\lbrack \frac{1 + \sqrt{37}}{2},\frac{1 + \sqrt{111}}{2} \right\rbrack
∵\because k is an integer,d iw.kkZad
∴k∈{−4,−3,4,5}\therefore k \in \{ - 4, - 3,4,5\}
number of possible integral values of k is

Source: Grass-G / M-25 Mathematics Paper 7 Jan.docx

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