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Home » Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x – 2y + 4 = 0.
Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x – 2y + 4 = 0.
CBSE | Circle | Class 11 | Exercise 24.1 | Maths | RD Sharma
Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x – 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x – 2y + 4 = 0.

solving the lines x + 3y = 0 and 2x – 7y = 0, we get the point of intersection to be (0, 0)

solving the lines x + y + 1 and x – 2y + 4 = 0, we get the point of intersection to be (-2, 1)

circle with centre (-2, 1) and passing through the point (0, 0).

Since, radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.

    \[\begin{array}{*{35}{l}} {{\left( x\text{ }-\text{ }p \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }q \right)}^{2}}~=\text{ }{{r}^{2}} \\ Where,\text{ }p\text{ }=\text{ }-2,\text{ }q\text{ }=\text{ }1 \\ {{\left( x\text{ }+\text{ }2 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }1 \right)}^{2}}~=\text{ }{{r}^{2}}~\ldots .\text{ }\left( 1 \right) \\ \end{array}\]

Equation (1) passes through (0, 0)

So,

    \[\begin{array}{*{35}{l}} {{\left( 0\text{ }+\text{ }2 \right)}^{2}}~+\text{ }{{\left( 0\text{ }-\text{ }1 \right)}^{2}}~=\text{ }{{r}^{2}} \\ 4\text{ }+\text{ }1\text{ }=\text{ }{{a}^{2}} \\ 5\text{ }=\text{ }{{r}^{2}} \\ r\text{ }=\text{ }\surd 5 \\ \end{array}\]

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2

By substitute the values in the above equation, we get

    \[\begin{array}{*{35}{l}} {{\left( x\text{ }\text{ }-\left( -2 \right) \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }1 \right)}^{2}}~=\text{ }{{\left( \surd 5 \right)}^{2}} \\ {{\left( x\text{ }+\text{ }2 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }1 \right)}^{2}}~=\text{ }5 \\ {{x}^{2}}~+\text{ }4x\text{ }+\text{ }4\text{ }+\text{ }{{y}^{2}}~-\text{ }2y\text{ }+\text{ }1\text{ }=\text{ }5 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~+\text{ }4x\text{ }-\text{ }2y\text{ }=\text{ }0 \\ \end{array}\]

∴ The equation of the circle is x2 + y2 + 4x – 2y = 0.

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