Answer:
<u>Let the center of the circle be A(x,y)</u>
<u>Let the center of the circle be A(x,y)Then PA=PQ</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2(x−6)2+(y+6)2=(x−3)2+(y+7)2</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2(x−6)2+(y+6)2=(x−3)2+(y+7)23x+y−7=0 --- (i)</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2(x−6)2+(y+6)2=(x−3)2+(y+7)23x+y−7=0 --- (i)AP=AR</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2(x−6)2+(y+6)2=(x−3)2+(y+7)23x+y−7=0 --- (i)AP=AR(AP)2=(AR)2</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2(x−6)2+(y+6)2=(x−3)2+(y+7)23x+y−7=0 --- (i)AP=AR(AP)2=(AR)2(x−6)2+(y+6)2=(x−3)2+(y−3)2</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2(x−6)2+(y+6)2=(x−3)2+(y+7)23x+y−7=0 --- (i)AP=AR(AP)2=(AR)2(x−6)2+(y+6)2=(x−3)2+(y−3)23x−9y−27=0 ---- (ii)</u>
<u>Let the center of the circle be A(x,y)Then PA=PQPA2=QA2(x−6)2+(y+6)2=(x−3)2+(y+7)23x+y−7=0 --- (i)AP=AR(AP)2=(AR)2(x−6)2+(y+6)2=(x−3)2+(y−3)23x−9y−27=0 ---- (ii)(ii) - (i), we get </u>
<u>y</u><u>=</u><u>-2</u>
<u>put in (i)</u>
<u>x</u><u>=</u><u>3</u>
<u>Therefore center of the circle is </u><u>(</u><u>3</u><u>,</u><u>-2</u><u>)</u>