#### Question

Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line *x *– 3*y* – 11 = 0.

#### Solution

Let the equation of the required circle be (*x* – *h*)^{2} + (*y* – *k*)^{2} = *r*^{2}.

Since the circle passes through points (2, 3) and (–1, 1),

(2 – *h*)^{2} + (3 – *k*)^{2} = *r*^{2} … (1)

(–1 – *h*)^{2} + (1 – *k*)^{2} = *r*^{2} … (2)

Since the centre (*h*, k) of the circle lies on line *x *– 3*y* – 11 = 0,

*h *– 3*k* = 11 … (3)

From equations (1) and (2), we obtain

(2 – *h*)^{2 }+ (3 – *k*)^{2} = (–1 – *h*)^{2} + (1 – *k*)^{2}

⇒ 4 – 4*h* + *h*^{2} + 9 – 6*k* + *k*^{2} = 1 + 2*h* + *h*^{2} + 1 – 2*k* + *k*^{2}

⇒ 4 – 4*h* + 9 – 6*k* = 1 + 2*h* + 1 – 2*k*

⇒ 6*h* + 4*k* = 11 … (4)

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Solution for question: Find the Equation of the Circle Passing Through the Points (2, 3) and (–1, 1) and Whose Centre is on the Line X – 3y – 11 = 0. concept: Concept of Circle. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts)