(iii)
Let the circle touches the x-axis at the point (a, 0) and y-axis at the point (0, a).
Then the centre of the circle is (a, a) and radius is a.
=> (x – p)2 + (y – q)2 = r2
substituting the values we get
(x – a)2 + (y – a)2 = a2 … (1)
equation (1) passes through P (2, 1)
substituting the values we get
![\[\begin{array}{*{35}{l}} {{\left( 2\text{ }-\text{ }a \right)}^{2}}~+\text{ }{{\left( 1\text{ }-\text{ }a \right)}^{2}}~=\text{ }{{a}^{2}} \\ 4\text{ }-\text{ }4a\text{ }+\text{ }{{a}^{2}}~+\text{ }1\text{ }-\text{ }2a\text{ }+\text{ }{{a}^{2}}~=\text{ }{{a}^{2}} \\ 5\text{ }-\text{ }6a\text{ }+\text{ }{{a}^{2}}~=\text{ }0 \\ \left( a\text{ }-\text{ }5 \right)\text{ }\left( a\text{ }-\text{ }1 \right)\text{ }=\text{ }0 \\ So,\text{ }a\text{ }=\text{ }5\text{ }or\text{ }1 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-04242a47f2dc0e74db81693830a52750_l3.png "Rendered by QuickLaTeX.com")
Case (i)
As, the centre at (5, 5) and having radius 5 units.
As, the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2
Now by substituting the values in the equation we get
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }-\text{ }5 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }5 \right)}^{2}}~=\text{ }{{5}^{2}} \\ {{x}^{2}}~-\text{ }10x\text{ }+\text{ }25\text{ }+\text{ }{{y}^{2}}~-\text{ }10y\text{ }+\text{ }25\text{ }=\text{ }25 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }10x\text{ }-\text{ }10y\text{ }+\text{ }25\text{ }=\text{ }0. \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9545dffba242cf0376bbb9dbd98ea638_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 – 10x – 10y + 25 = 0.
Case (ii)
Since, the centre at (1, 1) and having a radius 1 unit.
the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2
substituting the values in the equation we get
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }-\text{ }1 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }1 \right)}^{2}}~=\text{ }{{1}^{2}} \\ {{x}^{2}}~-\text{ }2x\text{ }+\text{ }1\text{ }+\text{ }{{y}^{2}}~-\text{ }2y\text{ }+\text{ }1\text{ }=\text{ }1 \\ {{x}^{2}}~+\text{ }{{y}^{2}}~-\text{ }2x\text{ }-\text{ }2y\text{ }+\text{ }1\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-66257ae4a7edb09c2c19da8673fea834_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 – 2x – 2y + 1 = 0.
(iv) Passing through the origin, radius 17 and ordinate of the centre is – 15.
Let us assume the abscissa as ‘a’
We have a circle with centre (a, – 15) and passing through the point (0, 0) and having radius 17.
We know that the radius of the circle is the distance between the centre and any point on the radius. So, we find the radius of the circle.
By using the distance formula,
![\[\begin{array}{*{35}{l}} {{17}^{2}}~=\text{ }{{a}^{2}}~+\text{ }{{\left( -15 \right)}^{2}} \\ 289\text{ }=\text{ }{{a}^{2}}~+\text{ }225 \\ {{a}^{2}}~=\text{ }64 \\ \left| a \right|\text{ }=\text{ }\surd 64 \\ \left| a \right|\text{ }=\text{ }8 \\ a\text{ }=\text{ }\pm 8\text{ }\ldots .\text{ }\left( 1 \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bb0881319239184600604d9d100b8d64_l3.png "Rendered by QuickLaTeX.com")
the centre at (±8, – 15) and having radius 17 units.
Since, the equation of the circle with centre (p, q) and having radius ‘r’ is given by: (x – p)2 + (y – q)2 = r2
Now by substituting the values in the equation, we get
![\[\begin{array}{*{35}{l}} {{\left( x\text{ }\pm \text{ }8 \right)}^{2}}~+\text{ }{{\left( y\text{ }-\text{ }15 \right)}^{2}}~=\text{ }{{17}^{2}} \\ {{x}^{2~}}\pm \text{ }16x\text{ }+\text{ }64\text{ }+\text{ }{{y}^{2}}~-\text{ }30y\text{ }+\text{ }225\text{ }=\text{ }289 \\ {{x}^{2}}~+\text{ }{{y}^{2~}}\pm \text{ }16x\text{ }-\text{ }30y\text{ }=\text{ }0. \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9bc61960c2bc8c6c73d40d5fbf51c977_l3.png "Rendered by QuickLaTeX.com")
∴ The equation of the circle is x2 + y2 ± 16x – 30y = 0.