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Exercise 5.1 · Q11

Q.Find centre and radius of the following circles.

(i) x2+(y+2)2=0x^2+(y+2)^2=0
(ii) x2+y2+6x−4y+4=0x^2+y^2+6x-4y+4=0
(iii) x2+y2−x+2y−3=0x^2+y^2-x+2y-3=0
(iv) 2x2+2y2−6x+4y+2=02x^2+2y^2-6x+4y+2=0
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Each part is a direct application of x2+y2+2gx+2fy+c=0⇒x^2+y^2+2gx+2fy+c=0 \Rightarrow centre (−g,−f)(-g,-f), radius g2+f2−c\sqrt{g^2+f^2-c}; part (iv) first needs dividing by the leading coefficient 22 to match this form.

Part (i). x2+(y+2)2=0x^2+(y+2)^2=0. This is already standard form (x−0)2+(y+2)2=02(x-0)^2+(y+2)^2=0^2: centre (0,−2)(0,-2), radius 00 — a degenerate point circle (the "locus" is the single point (0,−2)(0,-2)).

Part (ii). x2+y2+6x−4y+4=0x^2+y^2+6x-4y+4=0. 2g=6⇒g=32g=6\Rightarrow g=3; 2f=−4⇒f=−22f=-4\Rightarrow f=-2; c=4c=4. Centre (−3,2)(-3,2); radius =9+4−4=9=3=\sqrt{9+4-4}=\sqrt9=3.

Part (iii). x2+y2−x+2y−3=0x^2+y^2-x+2y-3=0. 2g=−1⇒g=−122g=-1\Rightarrow g=-\frac12; 2f=2⇒f=12f=2\Rightarrow f=1; c=−3c=-3. Centre (12,−1)\left(\frac12,-1\right); radius =14+1+3=174=172=\sqrt{\frac14+1+3}=\sqrt{\frac{17}4}=\frac{\sqrt{17}}2. …

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