The image of the circle x2+y2+16x−24y+183=0 in the line mirror 4x+7y+13=0, is

The equation of the given circle isx2+y2+16x−24y+183=0⇒x2+16x+64+y2−24y+144=25⇒(x+8)2+(y−12)2=52⇒{x−(−8)}2+(y−12)2=52Clearly, the centre of this circle is (- 8, 12) and radius= 5.The image of the circle (i) in the line mirror 4x+7y+13=0 is a ircle whose centre is the image of the point ( - 8, 12) in the linemirror 4x+7y+13=0 and radius same as that of the given circle.Let (h, k) be the image of (- 8, 12) in the line mirror4x+7y+13=0. The,h+84=k−127=−2(4×(−8)+7×12+13)42+72⇒h+84=k−127=−13065⇒h=−16, k=−2Hence, the centre of the required circle is (-16, - 2) and radiusequal to 5 units. So, its equation is(x+16)2+(y+2)2=52