Question

Find the equation of the circle whose center is on the linr y - x + 1 = and passing through the point (7,3) and its radius is 3​

Answer 1

Given information

• ⇀The circle whose center is on the linr y - x + 1 = 0 and passing through the point (7,3) and its radius is 3

What we need to calculate

• ⇀We have to find the equation of the circle

let's Calculate

Let (h,k) be the center of the circle, then

$\qquad \qquad \: \red{ \textsf{ \textbf{k - h + 1 = 0}}} \qquad \: \dots(1) \\ \\ \sf \: and \quad \: \sqrt{(h - {7)}^{2} + (k - {3)}^{2} } = 3 \\ \\ \sf \implies\: (h \: { - 7)}^{2} + (k { - 3)}^{2} = 9 \\ \\ \sf \implies \: {h}^{2} - 14h \: + 49 + {k }^{2} - 6k \: + 9 = 9 \\ \\ \red{\bf{ \implies \: {h}^{2} + {k}^{2} - 14h - 6k + 49 = 0} \: } \qquad \dots(2)$

From equation (1),

$\\ \qquad \qquad \qquad \textsf {\textbf{k = h - 1}} \qquad \qquad \dots(3)$

Putting this value of k in equation (2),

$\\ \qquad \sf \: {h}^{2} + {(h - 1)}^{2} - 14h - 6(h - 1) + 49 = 0 \\ \\ \sf \implies \: {h}^{2} + {h}^{2} - 2h + 1 - 14h - 6h + 6 + 49 = 0 \\ \\ \sf \implies \qquad \qquad \qquad \: 2 {h}^{2} - 22h + 56 = 0 \\ \\ \sf \implies \qquad \qquad \qquad \: {h}^{2} - 11h + 28 = 0 \\ \\ \sf \implies \qquad \qquad \qquad \: {h}^{2} - 4h - 7h + 28 = 0 \\ \\ \sf \implies \qquad \qquad \qquad \:(h - 4)(h - 7) = 0 \\ \\ \implies{\qquad \qquad \qquad \underline{\boxed{\textbf{ \textsf{ \:h = 4 ,7}}}}} \green \bigstar$

From equation (3)

$\qquad \qquad \qquad \qquad \bf k = 4 - 1,7 - 1 \\ \\ \implies \qquad \qquad \qquad \bf \: k= 3,6$

Therefore, center of the circle is (4,3) and (7,6) and radius is 3

Thus , equation of the circle is

$\qquad \qquad \: \sf \: {(x - 4)}^{2} + {(y - 3)}^{2} = {3}^{2} \\ \\ \sf \implies \qquad \qquad \: {x}^{2} - 8x + 16 + {y}^{2} - 6y + 9 = 9 \\ \\ \sf \implies \qquad \qquad \qquad \: {x}^{2} - {y}^{2} - 8x - 6y + 16 = 0 \\ \\ \bf \: and \: \qquad \qquad \sf \: {(x - 7)}^{2} + {(y - 6)}^{2} = {3}^{2} \\ \\ \sf \implies \qquad \: {x}^{2} - 14 + 49 + {y}^{2} - 12y + 36 = 9 \\ \\ \sf \implies \qquad \underline{ \boxed{{ \bf {x}^{2} + {y}^{2} - 14x - 12y + 76 = 0}}} \orange\bigstar$