Question -
Answer -
Let the equation of the requiredcircle be (x┬атАУ┬аh)2┬а+ (y┬атАУ┬аk)2┬а=┬аr2.
Sincethe circle passes through (0, 0),
(0тАУ┬аh)2┬а+ (0 тАУ┬аk)2┬а=┬аr2
тЗТ┬аh2┬а+┬аk2┬а=┬аr2
Theequation of the circle now becomes (x┬атАУ┬аh)2┬а+ (y┬атАУ┬аk)2┬а=┬аh2┬а+┬аk2.
Itis given that the circle makes intercepts┬аa┬аand┬аb┬аonthe coordinate axes. This means that the circle passes through points (a,0) and (0,┬аb). Therefore,
(a┬атАУ┬аh)2┬а+ (0 тАУ┬аk)2┬а=┬аh2┬а+┬аk2┬атАж(1)
(0тАУ┬аh)2┬а+ (b┬атАУ┬аk)2┬а=┬аh2┬а+┬аk2┬атАж (2)
Fromequation (1), we obtain
a2┬атАУ 2ah┬а+┬аh2┬а+┬аk2┬а=┬аh2┬а+┬аk2
тЗТ┬аa2┬атАУ 2ah┬а= 0
тЗТ┬аa(a┬атАУ2h)= 0
тЗТ┬аa┬а=0 or (a┬атАУ2h)= 0
However,┬аa┬атЙа0; hence, (a┬атАУ2h)= 0 тЗТ┬аh┬а=
.
Fromequation (2), we obtain
h2┬а+┬аb2┬атАУ2bk┬а+┬аk2┬а=┬аh2┬а+┬аk2
тЗТ┬аb2┬атАУ 2bk┬а= 0
тЗТ┬аb(b┬атАУ2k)= 0
тЗТ┬аb┬а=0 or(b┬атАУ2k)= 0
However,┬аb┬атЙа0; hence, (b┬атАУ2k)= 0 тЗТ┬аk┬а=┬а
.
Thus,the equation of the required circle is
