Let 0 \( \Large \left(\alpha ,\ \beta\right) \) and r be the centre and radius of the given circle respectively. Let M (h. k) be the middle point of the chord AB of length 2d Clearly, \( \Large \left(2d < 2r, or \ d < r\right) \). Since, M is the middle point of AB, therefore OM is perpendicular to AB. Therefore, \( \Large OA^{2}=AM^{2}+OM^{2} \) => \( \Large r^{2}=d^{2}+ \left(h-a\right)^{2}+ \left(k - \beta \right)^{2} \) Hence M lies on the circle, \( \Large \left(x- \alpha \right)^{2}+ \left(y- \beta \right)^{2} = r^{2} - d^{2} \) It has the same centre but less radius than the given circle.