A circle has its centre in the first quadrant and passes through (2,3). If this circle makes intercepts of length 3 and 4 respectively on x=2 and y=3, its equation is
Let the centre of circle be (h,k). Equation of circle will be x2+y2−2hx−2ky+c=0........(i) Circle passes through (2,3) ‌22+32−4h−6k+c=0 ⇒‌‌4h+6k−c=13 ‌ or ‌‌c=4h+6k−13 Equation of circle be x2+y2−2hx−2ky+4h+6k−13 This circle makes intercept of length 3 on the line x=2. |y1−y2
‌
‌
|=3, when x=2 is substituted in the equation of circle, where y1 and y2 are the two roots ‌4+y2−4h−2ky+4h+6k−13=0 ‌⇒y2−2ky+6k−9=0 ‌y1+y2=2k‌ and ‌y1y2=6k−9 ‌|y1−y2
in y1y2, we get y1y2 as 0] Now, similarly substitute y=3 ‌x2+9−2hx−6k+4h+6k−13=0 ⇒x2−2hx+4h−4=0 ⇒x1+x2=2h1,x1x2=4h−4 ∵‌‌|x1−x2
‌
‌
|=4 ‌√(x1+x2)2−4x1x2=4 ⇒4h2−16h+16=16⇒4h(h−4)=0 ‌h=4‌ and ‌h≠0 ∴ Equation of the required circle is x2+y2−8x−9y+30=0