\(L_1:4x+3y+2=0\)
\(L_2:3x−4y−11=0\)
Since circle C touches the line L2 at Q intersection point Q of L1 and L2, is (1, –2)
\(∵\) P lies of L1
\(∴\) \(P(x,\frac{−1}{3}(2+4x))\)
Now,
\(PQ=5⇒\) \((x−1)^2+(\frac{4x+2}{3}−2)^2=25\)
\(⇒(\)\(x−1)^2[1+\frac{16}{9}]=25\)
\(⇒ (x – 1)^2 = 9\)
\(⇒ x = 4, –2\)
\(∵\) The circle lies below the x-axis.
\(y = –6\)
\(P(4, –6)\)
Now distance of P from \(5x – 12y + 51 = 0\)
\(=\) \(\left| \frac{20 + 72 + 51}{13} \right| = \frac{143}{13} = 11\)
If the inverse point of the point \( (-1, 1) \) with respect to the circle \( x^2 + y^2 - 2x + 2y - 1 = 0 \) is \( (p, q) \), then \( p^2 + q^2 = \)
If the domain of the function \( f(x) = \frac{1}{\sqrt{3x + 10 - x^2}} + \frac{1}{\sqrt{x + |x|}} \) is \( (a, b) \), then \( (1 + a)^2 + b^2 \) is equal to:
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
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